FREE ALGEBRA HELP - SPECIAL PRODUCTS, EXAMPLES, EXERCISE USING FORMULAS

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Please study
Algebra Formulas before Free Algebra Help
if you have not already done so.
There we have listed out all the
Algebra Formulas that need to be
remembered.

Also Please study
Proofs of the first six Formulas

Proofs of the Last six Formulas

if you have not already done so.

Here we present Solved Examples
and Exercise problems
on application of those Formulas.

Example 1 of Free Algebra Help

Solve the following problems on Free Algebra Help.

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

Solution to the Example 1 on Free Algebra Help :
(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.

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Example 2 of Free Algebra Help

Solved Example 2 of Free Algebra Help :

Solve the following problems on Free Algebra Help.

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

Solution to the Example 2 on Free Algebra Help :
(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.

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Example 3 of Free Algebra Help

Solved Example 3 of Free Algebra Help :

Solve the following problems on Free Algebra Help.

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

Solution to the Example 3 on Free Algebra Help :
(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.

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Exercise : Free Algebra Help

Solve the following problems on Free Algebra Help.

  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
For Answers See at the bottom of the page.

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Answers to Exercise : Free Algebra Help

Answers to problems on Free Algebra Help :

  1. (i) 136 (ii) 1540
  2. (i) 23 (ii) 527 (iii) 110
  3. (i) 1004004 (ii) 1006012008

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