Let P = 64a^{3} - 336a^{2} + 588a - 343 We know 64 = 4 x 4 x 4 = 4^{3}; 336 = 7 x 48 = 7 x 3 x 16 = 3 x 4^{2} x 7; 588 = 7 x 84 = 7 x 7 x 12 = 7^{2} x 3 x 4 = 3 x 4 x 7^{2}; 343 = 7 x 49 = 7 x 7 x 7 = 7^{3} ∴ P = (4a)^{3} - 3(4a)^{2}(7) + 3(4a)(7)^{2} - 7^{3} This looks like (first term)^{3} - 3(first term)^{2}(second term) + 3(first term)(second term)^{2} - (second term)^{3} which is equal to (first term - second term)^{3} [See Formula 7], with (4a) in place of first term and 7 in place of second term. Applying this Formula here, we get P = (4a - 7)^{3}.

Thus Algebra Factoring of 64a^{3} - 336a^{2} + 588a - 343by using Algebra Formulas gave the factors as (4a - 7)^{3}. Ans.

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Let P = a^{6} + 54a^{3} + 729 We know 729 = 9 x 81 = 9 x 3 x 27 = 27 x 27 = 27^{2}; 54 = 2 x 27; Let t = a^{3}; then a^{6} = (a^{3})^{2} = t^{2} ∴ P = t^{2} + 2(27)t + (27)^{2} This looks like a^{2} + 2ab + b^{2} which is equal to (a + b)^{2} [ See Formula 1], with (t) in place of a and 27 in place of b ∴ P = (t + 27)^{2} (t + 27) = a^{3} + 3^{3} = (a + 3){a^{2} - a(3) + 3^{2}} [See Formula 4] = (a + 3)(a^{2} - 3a + 9) ∴ P = (t + 27)^{2} = {(a + 3)(a^{2} - 3a + 9)}^{2}= (a + 3)^{2}(a^{2} - 3a + 9)^{2}.

Thus Algebra Factoring of a^{6} + 54a^{3} + 729by using Algebra Formulas gave the factors as (a + 3)^{2}(a^{2} - 3a + 9)^{2}. Ans.

Example 3 : Factoring In Algebra

Solved Example 3 of Factoring In Algebra :

If x^{3} + y^{3} + z^{3} = 3xyz, show that either x + y + z = 0 or x = y = z

solution to Example 3 of Factoring In Algebra :

By data x^{3} + y^{3} + z^{3} = 3xyz ⇒ x^{3} + y^{3} + z^{3} - 3xyz = 0 We have a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca) [See Formula 9] Applying this here, we get (x + y + z)(x^{2} + y^{2} + z^{2} - xy - yz - zx) = 0 ⇒ either (x + y + z) = 0 or (x^{2} + y^{2} + z^{2} - xy - yz - zx) = 0 Let P = (x^{2} + y^{2} + z^{2} - xy - yz - zx) Multiplying both sides with 2, we get 2P = (2x^{2} + 2y^{2} + 2z^{2} - 2xy - 2yz - 2zx) Writng 2x^{2} as x^{2} + x^{2}, 2y^{2} as y^{2} + y^{2} and 2z^{2} as z^{2} + z^{2} and regrouping the terms, we get 2P = {(x^{2} + y^{2} - 2xy) + (y^{2} + z^{2} - 2yz) + (z^{2} + x^{2} - 2zx)} = {(x - y)^{2} + (y - z)^{2} + (z - x)^{2}} P = 0 ⇒ 2P = 2 x 0 = 0 ⇒ {(x - y)^{2} + (y - z)^{2} + (z - x)^{2}} = 0 SUM OF SQUARES OF THREE TERMS IS ZERO ⇒ EACH TERM IS ZERO. ∴ (x - y) = 0, (y - z) = 0 and (z - x) = 0 ⇒ x = y, y = z and z = x ⇒ x = y = z Thus, If x^{3} + y^{3} + z^{3} = 3xyz, then either x + y + z = 0 or x = y = z.

Thus, by Algebra Factoring of x^{3} + y^{3} + z^{3} - 3xyz, by using Algebra Formulas, we have proved that either x + y + z = 0 or x = y = z if x^{3} + y^{3} + z^{3} = 3xyz.

Let P = (4x - 3y)^{3} + (y - 3x)^{3} + (2y - x)^{3} Let A = (4x - 3y), B = (y - 3x) and C = (2y - x) Then A + B + C = (4x - 3y) + (y - 3x) + (2y - x) = 4x - 4x + 3y - 3y = 0 We can prove If A + B + C = 0, A^{3} + B^{3} + C^{3} = 3ABC For proof, see the Example 12 of
Algebra Formulas. ∴ A + B + C = 0 ⇒ A^{3} + B^{3} + C^{3} = 3ABC ∴ P = (4x - 3y)^{3} + (y - 3x)^{3} + (2y - x)^{3}= A^{3} + B^{3} + C^{3} and also A + B + C = 0 ∴ P = 3ABC = 3(4x - 3y)(y - 3x)(2y - x).

Thus Algebra Factoring of (4x - 3y)^{3} + (y - 3x)^{3} + (2y - x)^{3}by using Algebra Formulas gave the factors as 3(4x - 3y)(y - 3x)(2y - x). Ans.

Exercise : Factoring In Algebra

Solve the following problems in Factoring In Algebra.

Factorize 125x^{3} - 75x^{2} + 15x - 1

Factorize 64x^{6} + 16x^{3} + 1

Factorize 729a^{6} - 64b^{6}

Factorize 27x^{3} + 64y^{3} + 36xy -1

Factorize (x - y)^{3} + (y - z)^{3} + (z - x)^{3}

For Answers See at the bottom of the Page.

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