HOW TO DO ALGEBRA - SPECIAL PRODUCTS, EXAMPLES, EXERCISE, USING FORMULAS

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

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

Example 1 of How To Do Algebra

Find the quotient (27x^{3} - 64)÷(9x^{2} + 12x + 16) without actual division.

Solution to Example 1 of How To Do Algebra : Let P = (27x^{3} - 64)÷(9x^{2} + 12x + 16) We know 27 = 3 x 3 x 3 = 3^{3} ⇒27x^{3} = 3^{3}x^{3} = (3x)^{3} 64 = 4 x 4 x 4 = 4^{3} ∴ Numerator of P = (27x^{3} - 64) = (3x)^{3} - 4^{3} This looks like (a^{3} - b^{3}) with (3x) in place of a and (4) in place of b Denominator of P = (9x^{2} + 12x + 16) = (3x)^{2} + (3x)(4) + 4^{2} This looks like (a^{2} + ab + b^{2}) with (3x) in place of a and (4) in place of b We have (a^{3} - b^{3}) = (a - b)(a^{2} + ab + b^{2}) (See Formula 5) ⇒ (a^{3} - b^{3})÷(a^{2} + ab + b^{2}) = (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.

Solution to Example 2 of How To Do Algebra : Let P = 4x^{2} + 9y^{2} + 16z^{2} = (2x)^{2} + (3y)^{2} + (4z)^{2} This looks like a^{2} + b^{2} + c^{2} 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} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca (See Formula 8) Putting (-c) in place of c, we get (a + b - c)^{2} = a^{2} + b^{2} + (-c)^{2} + 2ab + 2b(-c) + 2(-c)a (See Formula 8) = a^{2} + b^{2} + c^{2} + 2ab - 2bc - 2ca ⇒ a^{2} + b^{2} + c^{2}= (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.

Find the quotient (x^{3} + 27y^{3} + 8z^{3} - 18xyz)÷(x + 3y + 2z) without actual division.

Solution to Example 3 of How To Do Algebra : We have a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca) [See Formula 9] ⇒ (a^{3} + b^{3} + c^{3} - 3abc)÷(a + b + c) = (a^{2} + b^{2} + c^{2} - ab - bc - ca).....(i) Let P = (x^{3} + 27y^{3} + 8z^{3} - 18xyz)÷(x + 3y + 2z) = {x^{3} + (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 = {x^{3} + (3y)^{3} + (2z)^{3} - 3(x)(3y)(2z)}÷(x + 3y + 2z) = {(x)^{2} + (3y)^{2} + (2z)^{2} - (x)(3y) - (3y)(2z) - (2z)(x)} = {x^{2} + 9y^{2} + 4z^{2} - 3xy - 6yz - 2zx} Ans.

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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 to Example 4 of How To Do Algebra : (i) Let P = (x + 1)(x - 3) We have (x + a)(x + b) = x^{2} + x(a + b) + ab [See Formula 10] Comparing P with this Formula, a = 1 and b = -3 ∴ P = (x + 1)(x - 3) = x^{2} + x{1 + (-3)} + {(1)(-3)} = x^{2} + x{-2} + {(-3)} = x^{2} - 2x - 3 Ans.

(ii) Let P = (2x + 5)(5x - 3) We have(ax + b)(cx + d) = acx^{2} + 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)x^{2} + x{(2)(-3) + (5)(5)} + (5)(-3) = 10x^{2} +19x - 15. Ans.

Example 5 of How To Do Algebra

Solved Example 5 on How To Do Algebra :

If l + m + n = 0, prove that l^{3} + m^{3} + n^{3} = 3lmn

Solution to Example 5 of How To Do Algebra : l + m + n = 0 ⇒ l + m = -n .........(i) Cubing both sides, we get (l + m)^{3} = (-n)^{3} ⇒ l^{3} + m^{3} + 3lm(l + m) = -n^{3} [See Formula 6] Using the value of (l + m) from (i), we get l^{3} + m^{3} + 3lm(-n) = -n^{3} ⇒ l^{3} + m^{3} - 3lmn = -n^{3} ⇒ l^{3} + m^{3} + n^{3} = 3lmn (Proved.)

Exercise : How To Do Algebra

Solve the following problems on How To Do Algebra

Find the quotient (8x^{3} - 343)÷(2x - 7) without actual division.

If x + 2y + 3z = 12 and x^{2} + 4y^{2} + 9z^{2} = 44, find 2xy + 6yz + 3zx

Find the quotient (l^{3} + 27m^{3} + 343n^{3} - 63lmn)÷(l^{2} + 9m^{2} + 49n^{2} - 3lm - 21mn - 7nl) without actual division.

Find the products (i) (x - 9)(x - 7) (ii) (5x + 2)(7x + 5) (iii) (x - 3)(x - 8)(x + 6) using the Algebra Formulas.

If p + q = r, prove that p^{3} + q^{3} + 3pqr = r^{3}

For Answers See at the bottom of the page.

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