ALGEBRA FORMULAS - SPECIAL PRODUCTS, EXAMPLES, EXERCISE, LINKS TO OTHER FORMULAS

Please study Polynomials before Algebra Formulas,
if you have not already done so.

There, we discussed definition, simplified form, addition,
subtraction, multiplication etc. of Polynomials.

That knowledge is a prerequisite here.

Here, We deal with Some Special Products in Polynomials.

Certain products of Polynomials occur more often
in Algebra. They are to be considered specially.

These are to be remembered as Formulas in Algebra.

Remembering these formulas in Algebra is as important
as remembering multiplication tables in Arithmetic.

We give a list of these Formulas and Apply
them to solve a Number of problems.

We give Links to other Formulas in Algebra.

Here is the list of Algebra Formulas in
Polynomials which are very useful in Algebra.

Algebra Formulas in Polynomials :

Algebra Formula 1 in Polynomials:

Square of Sum of Two Terms:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

(a + b)(a - b) = a² - b²

(a + b)(a² - ab + b²) = a³ + b³

(a - b)(a² + ab + b²) = a³ - b³

(a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + 3ab(a + b) + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³ = a³ - 3ab(a - b) - b³

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

(a + b + c)(a² + b² + c² - ab - bc - ca) = a³ + b³ + c³ - 3abc

(x + a)(x + b) = x² + x(a + b) + ab

(ax + b)(cx + d) = acx² + x(ad + bc) + bd

(x + a)(x + b)(x + c) = x³ + x²(a + b + c) + x(ab + bc + ca) + abc

Here a, b, c, d, x are all real numbers.

Each of the letters in fact represent a TERM.

e.g. The above Algebra Formula 1 can be stated as
(First term + Second term)²
= (First term)² + 2(First term)(Second term) + (Second term)²

Similarly in other Algebra Formulas also,
we can replace each of the letters by a TERM.

Proofs of Algebra Formulas in Polynomials :

Let us prove each of the Algebra formulas.

Proof of Algebra Formula 1:

To Prove:
(a + b)² = a² + 2ab + b²

L.H.S. = (a + b)² = (a + b)(a + b) = a(a + b) + b(a + b)
= a.a + a.b + b.a + b.b = a² + (ab + ab) + b²
= a² + 2ab + b² = R.H.S. (Proved.)

Proof of Algebra Formula 2:

To Prove:
(a - b)² = a² - 2ab + b²

Here we can directly prove as in Formula 1 above.
(I leave this as exercise for the reader.)

or, we can make use of Formula 1 to prove this.

Replacing b in Formula 1 by (-b), we get
{a + (-b)}² = a² + 2a(-b) + (-b)²
⇒ (a - b)² = a² - 2ab + b² (Proved.)

Proof of Algebra Formula 3:

To Prove:
(a + b)(a - b) = a² - b²

L.H.S. = (a + b)(a - b) = a(a - b) + b(a - b)
= a.a - a.b + b.a - b.b
= a² + (-ab + ab) - b² = a² - b² = R.H.S. (Proved.)

Proof of Algebra Formula 4:

To Prove:
(a + b)(a² - ab + b²) = a³ + b³

L.H.S. = (a + b)(a² - ab + b²)
= a(a² - ab + b²) + b(a² - ab + b²)

= a(a²) - a(ab) + a(b²) + b(a²) - b(ab) + b(b²)
= a³ - a²b + ab² + ba² - b²a + b³

= a³ + (- a²b + ba²) + (ab² - b²a) + b³
The terms in brackets are same and cancel each other.
∴ L.H.S. = a³ + b³ = R.H.S. (Proved.)

Proof of Algebra Formula 5:

To Prove:
(a - b)(a² + ab + b²) = a³ - b³

Here we can directly prove as in Formula 4 above.
(I leave this as exercise for the reader.)
or, we can make use of Formula 4 to prove this.

Replacing b in Formula 4 by (-b), we get
{a + (-b)}{a² - a(-b) + (-b)²} = a³ + (-b)³
⇒ (a - b)(a² + ab + b²) = a³ - b³ = R.H.S. (Proved.)

Proof of Algebra Formula 6:

To Prove:
(a + b)³ = a³ + 3a²b + 3ab² + b³

L.H.S. = (a + b)³ = (a + b)(a + b)²
Using Formula 1 here, we get
L.H.S. = (a + b)(a² + 2ab + b²)
= a(a² + 2ab + b²) + b(a² + 2ab + b²)
= a(a²) + a(2ab) + a(b²) + b(a²) + b(2ab) + b(b²)
= a³ + 2a²b + ab² + ba² + 2ab² + b³
= a³ + (2a²b + ba²) + (ab² + 2ab²) + b³
= a³ + 3a²b + 3ab² + b³ = R.H.S. (Proved.)

Proof of Algebra Formula 7:

To Prove:
(a - b)³ = a³ - 3a²b + 3ab² - b³

Here we can prove as in Formula 6 above.
(I leave this as exercise for the reader.)

or, we can make use of Formula 6 to prove this.

Replacing b in Formula 6 by (-b), we get
{a + (-b)}³ = a³ + 3a²(-b) + 3a(-b)² + (-b)³
⇒ (a - b)³ = a³ - 3a²b + 3ab² - b³ = R.H.S. (Proved.)

Proof of Algebra Formula 8:

To Prove:
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

L.H.S. = (a + b + c)² = { (a + b) + c }²
Applying Formula1 here,
treating (a + b) as first term and c as second term, we get
L.H.S. = (a + b)² + 2(a + b)c + c²
= a² + 2ab + b² + 2(ac + bc) + c²
= a² + b² + c² + 2ab + 2bc + 2ca = R.H.S. (Proved.)

Proof of Algebra Formula 9:

To Prove:
(a + b + c)(a² + b² + c² - ab - bc - ca) = a³ + b³ + c³ - 3abc

L.H.S. = a(a² + b² + c² - ab - bc - ca) + b(a² + b² + c² - ab - bc - ca)
+ c(a² + b² + c² - ab - bc - ca)
= (a.a² + a.b² + a.c² - a.ab - a.bc - a.ca)
+ (b.a² + b.b² + b.c² - b.ab - b.bc - b.ca)
+ (c.a² + c.b² + c.c² - c.ab - c.bc - c.ca)
= (a³ + ab² + ac² - a²b - abc - a²c)
+ (ba² + b³ + bc² - ab² - b²c - abc) + (ca² + cb² + c³ - abc - bc² - c²a)
= (a³ + b³ + c³ - abc - abc - abc)
+ (ab² - ab²)
+ (ac² - c²a)
+ ( - a²c + ca²)
Cancelling the same terms with opposite signs, we get
L.H.S. = (a³ + b³ + c³ - 3abc) = R.H.S. (Proved.)

Proof of Algebra Formula 10:

To Prove:
(x + a)(x + b) = x² + x(a + b) + ab

L.H.S. = (x + a)(x + b) = x(x + b) + a(x + b) = x.x + x.b + a.x + a.b
= x² + x(a + b) + ab = R.H.S. (Proved.)

Proof of Algebra Formula 11:

To Prove:
(ax + b)(cx + d) = acx² + x(ad + bc) + bd

L.H.S. = (ax + b)(cx + d) = ax(cx + d) + b(cx + d)
= ax.cx + ax.d + b.cx + b.d
= acx² + x(ad + bc) + bd = R.H.S. (Proved.)

Proof of Algebra Formula 12:

To Prove:
(x + a)(x + b)(x + c) = x³ + x²(a + b + c) + x(ab + bc + ca) + abc

L.H.S. = (x + a)(x + b)(x + c) = (x + a){(x + b)(x + c)}
Applying Formula 10 here, we get
L.H.S. = (x + a){x² + x(b + c) + bc}
= x{x² + x(b + c) + bc} + a{x² + x(b + c) + bc}
= x.x² + x.x(b + c) + x.bc + a.x² + a.x(b + c) + a.bc
= x³ + x²(b + c) + xbc + ax² + x(ab + ac) + a.bc
= x³ + x²(a + b + c) + x(ab + bc + ca) + abc = R.H.S. (Proved.)

Example 1 of Algebra Formulas

If 3x + 4y = 12 and xy = 2, find (i) 9x² + 16y² (ii) 27x³ + 64y³

Solution:
(i) Let P = 9x² + 16y² = (3x)² + (4y)²
This looks like a² + b² with 3x in place of a and 4y in place of b
So the problem becomes finding a² + b²
given (a + b) and ab{= (3x)(4y) = 12xy = 12(2)}
We know a² + 2ab + b² = (a + b)² (See Formula 1)
⇒ a² + b² = (a + b)² - 2ab
Applying this here to find P, we get
P = (3x)² + (4y)² = (3x + 4y)² - 2(3x)(4y) = (3x + 4y)² - 24xy
By data, 3x + 4y = 12 and xy = 2
Using these values here to find P, we get
P = (12)² - 24(2) = 144 - 48 = 96. Ans.

(ii) Let P = 27x³ + 64y³
we know 27 = 3 x 3 x 3 = 3³ ⇒ 27x³ = 3³x³ = (3x)³;
64 = 4 x 4 x 4 = 4³ ⇒ 64y³ = 4³y³ = (4y)³;
∴ P = (3x)³ + (4y)³
This looks like a³ + b³ with 3x in place of a and 4y in place of b
So the problem becomes finding a³ + b³
given (a + b) and ab {= (3x)(4y) = 12xy = 12(2)}
We know a³ + 3ab(a + b) + b³ = (a + b)³ (See Formula 6)
⇒ a³ + b³ = (a + b)³ - 3ab(a + b)
Applying this here to find P, we get
P = (3x)³ + (4y)³ = (3x + 4y)³ - 3(3x)(4y)(3x + 4y)
= (3x + 4y)³ - 36xy(3x + 4y)
By data, 3x + 4y = 12 and xy = 2
Using these values here to find P, we get
P = (12)³ - 36(2)(12) = 144 x 12 - 144 x 6
= 144(12 - 6) = 144 x 6 = 864. Ans.

Example 2 of Algebra Formulas

If (x - 1⁄x) = 4, find (i) x² + 1⁄x² (ii) x⁴ + 1⁄x⁴ (iii) x³ - 1⁄x³

Solution:
(i) Let P = x² + 1⁄x² = x² + (1⁄x)²
This looks like a² + b² with x in place of a and (1⁄x) in place of b
So the problem becomes finding a² + b²
given (a - b) and ab {= (x)(1⁄x) = 1}
We know a² - 2ab + b² = (a - b)² (See Formula 2)
⇒ a² + b² = (a - b)² + 2ab
Applying this here to find P, we get
P = x² + (1⁄x)² = (x - 1⁄x)² + 2(x)(1⁄x)
= (x - 1⁄x)² + 2(1) = (x - 1⁄x)² + 2
By data, (x - 1⁄x) = 4
Using this value here to find P, we get
P = (4)² + 2 = 16 + 2 = 18
∴ x² + 1⁄x² = 18. Ans.

(ii) Let P = x⁴ + 1⁄x⁴ = (x²)² + (1⁄x²)²
This looks like a² + b² with (x²) in place of a and (1⁄x²) in place of b
So the problem becomes finding a² + b²
given (a + b) { = x² + 1⁄x² = 18 from (i) } and ab {= (x²)(1⁄x²) = 1}
We know a² + 2ab + b² = (a + b)² (See Formula 1)
⇒ a² + b² = (a + b)² - 2ab
Applying this here to find P, we get
P = (x²)² + (1⁄x²)² = (x² + 1⁄x²)² - 2(x²)(1⁄x²)
= (x² + 1⁄x²)² - 2(1) = (x² + 1⁄x²)² - 2
From the result found in (i), (x² + 1⁄x²) = 18.
Using this value here to find P, we get
P = (18)² - 2 = 324 - 2 = 322
∴ x⁴ + 1⁄x⁴ = 322. Ans.

(iii) Let P = x³ - 1⁄x³ = x³ - (1⁄x)³
This looks like a³ - b³ with x in place of a and (1⁄x) in place of b
So the problem becomes finding a³ - b³
given (a - b) { = (x - 1⁄x) = 4 by data} and ab {= (x)(1⁄x) = 1}
We know a³ - 3ab(a - b) - b³ = (a - b)³ (See Formula 7)
⇒ a³ - b³ = (a - b)³ + 3ab(a - b) Applying this here to find P, we get
P = (x)³ - (1⁄x)³ = (x - 1⁄x)³ + 3(x)(1⁄x)(x - 1⁄x)
= (x - 1⁄x)³ + 3(1)(x - 1⁄x)
By data, x - 1⁄x= 4
Using these values here to find P, we get
P = (4)³ + 3(4) = 64 + 12 = 76. Ans.

Example 3 of Algebra Formulas

Find (i) 997² (ii) 997³ using Algebra Formulas

Solution:
(i) Let P = 997²
997 is nearer to 1000. We can write 997 = 1000 - 3.
∴ P = (1000 - 3)²;
This looks like (a - b)² with (1000) in place of a and (3) in place of b
We have (a - b)² = a² - 2ab + b² (See Formula 2)
∴ P = (1000 - 3)² = (1000)² - 2(1000)(3) + (3)² = 1000000 - 6000 + 9
= 1000(1000 - 6) + 9 = 1000(994) + 9 = 994000 + 9 = 994009. Ans.

(ii) Let P = 997³ = (1000 - 3)³
This looks like (a - b)³ with (1000) in place of a and (3) in place of b
We have (a - b)³ = a³ - 3a²b + 3ab² - b³ (See Formula 7)
∴ P = (1000 - 3)³ = (1000)³ - 3(1000)²(3) + 3(1000)(3)² - (3)³
= (1000) {(1000)² - 9000 + 27} - 27
= (1000) [(1000){(1000) - 9} + 27] - 27
= (1000) [(1000){991} + 27] - 27 = (1000) [{991000} + 27] - 27
= (1000) [{991027}] - 27 = 991027000 - 27 = 991026973. Ans.

Example 4 of Algebra Formulas

Find (i) (100.5)² (ii) (100.5)³ using Algebra Formulas

Solution:
(i) Let P = (100.5)² = (100 + 0.5)²
This looks like (a + b)² with (100) in place of a and (0.5) in place of b
We have (a + b)² = a² + 2ab + b² (See Formula 1)
∴ P = (100 + 0.5)² = (100)² + 2(100)(0.5) + (0.5)²
= 10000 + 100 + 0.25 = 100(100 + 1) +0.25
= 10100 + 0.25 = 10100.25.Ans.

(ii) Let P = (100.5)³ = (100 + 0.5)³
This looks like (a + b)³
with (100) in place of a and (0.5) in place of b
We have (a + b)³ = a³ + 3a²b + 3ab² + b³ (See Formula 7)
∴ P = (100 + 0.5)³ = 100³ + 3(100)²(0.5) + 3(100)(0.5)² + (0.5)³
= (100)² (100 + 1.5) + 75 + 0.125 = (101.5)10000 + 75.125
= 1015000 + 75.125 = 1015075.125 Ans.

Example 5 of Algebra Formulas

Find (2x - 3y)(2x + 3y)(4x² + 9y²) and verify the result for x = -1, y = 1.

Solution:
Let P = (2x - 3y)(2x + 3y)(4x² + 9y²)
(2x - 3y)(2x + 3y) looks like (a - b)(a + b) with 2x in place of a and 3y in place of b
We have (a - b)(a + b) = (a + b)(a - b) = a² - b² (See Formula 3)
∴ (2x - 3y)(2x + 3y) = (2x)² - (3y)² = (4x² - 9y²)
Using this in P, we get
P = (4x² - 9y²)(4x² + 9y²)
This again looks like (a - b)(a + b)
with 4x² in place of a and 9y² in place of b
So, Applying Formula 3 again, we get
P = (4x²)² - (9y²)² = 16x⁴ - 81y⁴. Ans.

Verifying the result for x = -1, y = 1 :

We have (2x - 3y)(2x + 3y)(4x² + 9y²) = (16x⁴ - 81y⁴)
For x = -1, y = 1, L.H.S. = (2x - 3y)(2x + 3y)(4x² + 9y²)
= {2(-1) - 3(1)}{2(-1) + 3(1)}{4(-1)² + 9(1)²)
= {-5}{1}{4 + 9} = -5 x 13 = -65
For x = -1, y = 1, R.H.S. = {16(-1)⁴ - 81(1)⁴)} = 16 - 81 = -65
∴ L.H.S. = R.H.S. (Verified.)

Example 6 of Algebra Formulas

Find 109 x 91 using Algebra Formula

Solution:
As you can see, the given product is
product of sum and difference of two numbers,
to which we can apply Formula 3.
Let P = 109 x 91 = (100 + 9)(100 - 9)
We have (a + b)(a - b) = a² - b² (See Formula 3)
∴ P = (100 + 9)(100 - 9) = (100)² - (9)² = 10000 - 81 = 9919. Ans.

Example 7 of Algebra Formulas

Find by what expression we should multiply (16x² - 20xy + 25y²)
to get the result as sum of two cubes.

Solution:
Let P = (16x² - 20xy + 25y²) = (4x)² - (4x)(5y) + (5y)²
This looks like (a² - ab + b²) with (4x) in place of a and (5y) in place of b
We have (a + b)(a² - ab + b²) = a³ + b³ (See Formula 4)
⇒ (a² - ab + b²) is to be multiplied with (a + b)
to get (a³ + b³) which is the sum of two cubes.
∴ P is to be multiplied with (4x + 5y)
to get the result as sum of two cubes. Ans.

Example 8 of Algebra Formulas

Find the quotient (27x³ - 64)÷(9x² + 12x + 16) without actual division.

Solution:
Let P = (27x³ - 64)÷(9x² + 12x + 16)
We know 27 = 3 x 3 x 3 = 3³ ⇒27x³ = 3³x³ = (3x)³
64 = 4 x 4 x 4 = 4³
∴ Numerator of P = (27x³ - 64) = (3x)³ - 4³
This looks like (a³ - b³) with (3x) in place of a and (4) in place of b
Denominator of P = (9x² + 12x + 16) = (3x)² + (3x)(4) + 4²
This looks like (a² + ab + b²) with (3x) in place of a and (4) in place of b
We have (a³ - b³) = (a - b)(a² + ab + b²) (See Formula 5)
⇒ (a³ - b³)÷(a² + ab + b²) = (a - b)
P is like the L.H.S. of this with (3x) in place of a and (4) in place of b
∴ P = (3x - 4) Ans.

Example 9 of Algebra Formulas

If 2x + 3y - 4z = 10 and 3xy - 6yz - 4zx = 15, find 4x² + 9y² + 16z².

Solution:
Let P = 4x² + 9y² + 16z² = (2x)² + (3y)² + (4z)²
This looks like a² + b² + c²
with (2x) in place of a, (3y) in place of b and (4z) in place of c
Then the given data looks like a + b - c = 10 and (1⁄2 )(ab - bc - ca) = 15
We have (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca (See Formula 8)
Putting (-c) in place of c, we get
(a + b - c)² = a² + b² + (-c)² + 2ab + 2b(-c) + 2(-c)a (See Formula 8)
= a² + b² + c² + 2ab - 2bc - 2ca
⇒ a² + b² + c² = (a + b - c)² - 2(ab - bc - ca)
Applying this to P, we get
P = (2x)² + (3y)² + (4z)²
= (2x + 3y - 4z)² - 2{(2x)(3y) - (3y)(4z) - (4z)(2x)}
= (2x + 3y - 4z)² - 2 x 2{(x)(3y) - (3y)(2z) - (4z)(x)}
= (2x + 3y - 4z)² - 4{(3xy) - (6yz) - (4zx)}
By data, 2x + 3y - 4z = 10 and 3xy - 6yz - 4zx = 15
Using these in P, we get
P = (10)² - 4(15) = 100 - 60 = 40. Ans.

Example 10 of Algebra Formulas

Find the quotient
(x³ + 27y³ + 8z³ - 18xyz)÷(x + 3y + 2z)
without actual division.

Solution:
We have a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca) [See Formula 9]
⇒ (a³ + b³ + c³ - 3abc)÷(a + b + c) = (a² + b² + c² - ab - bc - ca).....(i)
Let P = (x³ + 27y³ + 8z³ - 18xyz)÷(x + 3y + 2z)
= {x³ + (3y)³ + (2z)³ - 3(x)(3y)(2z)}÷(x + 3y + 2z)
This looks like Equation (i)
with x in place of a, 3y in place of b and 2z in place of c
Applying Equation (i) here, we get
P = {x³ + (3y)³ + (2z)³ - 3(x)(3y)(2z)}÷(x + 3y + 2z)
= {(x)² + (3y)² + (2z)² - (x)(3y) - (3y)(2z) - (2z)(x)}
= {x² + 9y² + 4z² - 3xy - 6yz - 2zx} Ans.

Example 11 of Algebra Formulas

Find the products
(i) (x + 1)(x - 3) (ii) (2x + 5)(5x - 3) (iii) (x + 2)(x + 4)(x + 5)
using the Algebra Formulas.

Solution:
(i) Let P = (x + 1)(x - 3)
We have (x + a)(x + b) = x² + x(a + b) + ab [See Formula 10]
Comparing P with this Formula, a = 1 and b = -3
∴ P = (x + 1)(x - 3) = x² + x{1 + (-3)} + {(1)(-3)}
= x² + x{-2} + {(-3)} = x² - 2x - 3 Ans.

(ii) Let P = (2x + 5)(5x - 3)
We have (ax + b)(cx + d) = acx² + x(ad + bc) + bd
Comparing P with this Formula, a = 2, b = 5, c = 5 and d = -3
∴ P = (2x + 5)(5x - 3) = (2)(5)x² + x{(2)(-3) + (5)(5)} + (5)(-3) = 10x² +19x - 15. Ans.

Example 12 of Algebra Formulas

If l + m + n = 0, prove that l³ + m³ + n³ = 3lmn

Solution:
l + m + n = 0 ⇒ l + m = -n .........(i)
Cubing both sides, we get
(l + m)³ = (-n)³
⇒ l³ + m³ + 3lm(l + m) = -n³ [See Formula 6]
Using the value of (l + m) from (i), we get
l³ + m³ + 3lm(-n) = -n³
⇒ l³ + m³ - 3lmn = -n³
⇒ l³ + m³ + n³ = 3lmn (Proved.)

Other Algebra Formulas

The Following Links take you to other Algebra Formulas.

Formulas in Exponents

Formulas in Logarithms

Quadratic Formula

Exercise 1 : Algebra Formulas

1. If 2x - 3y = 10 and xy = 3, find (i) 4x² + 9y² (ii) 8x³ - 27y³
2. If x + 1⁄x = 5, find (i)x² + 1⁄x² (ii) x⁴ + 1⁄x⁴ (iii) x³ + 1⁄x³
3. Find (i) 1002² (ii) 1002³ using Algebra Formulas
4. Find (i) (99.5)² (ii) (99.5)³ using Algebra Formulas
5. Find (p - 4q)(p + 4q)(p² + 16q²)
   and verify the result for p = 2, q = 1.
6. Find (30.5) x (29.5) using Algebra Formula
7. Find the product (2x + 3y)(4x² - 6xy + 9y²)
   without actual multiplication
8. Find the quotient (8x³ - 343)÷(2x - 7) without actual division.
9. If x + 2y + 3z = 12 and x² + 4y² + 9z² = 44, find 2xy + 6yz + 3zx
10. Find the quotient
    (l³ + 27m³ + 343n³ - 63lmn)÷(l² + 9m² + 49n² - 3lm - 21mn - 7nl)
    without actual division.
11. Find the products
    (i) (x - 9)(x - 7) (ii) (5x + 2)(7x + 5) (iii) (x - 3)(x - 8)(x + 6)
    using the Algebra Formulas.
12. If p + q = r, prove that p³ + q³ + 3pqr = r³