We know a^3 b^3 =(a b) (a^2 +ab +b^2) a^6 b^6 = (a^2)^3 (b^2)^3 =(a^2 b^2) ((a^2)^2 + (a^2) × (b^2) +(b^2)^2) = (a^2 b^2) (a^4 + a^2b^2 +b^4) = (a+b ...

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Byju's Answer

Standard VIII

Mathematics

Factorisation Using Algebraic Identities

Factorise: a6...

Question

Factorise:

$a^{6} - b^{6}$

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Solution

$W e k n o w a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) a^{6} - b^{6} = (a^{2})^{3} - (b^{2})^{3} = (a^{2} - b^{2}) ((a^{2})^{2} + (a^{2}) \times (b^{2}) + (b^{2})^{2}) = (a^{2} - b^{2}) (a^{4} + a^{2} b^{2} + b^{4}) = (a + b) (a - b) (a^{4} + a^{2} b^{2} + b^{4})$

Suggest Corrections

Factorise: a6−b6

We know a3−b3=(a−b)(a2+ab+b2)a6−b6=(a2)3−(b2)3=(a2−b2)((a2)2+(a2)×(b2)+(b2)2)=(a2−b2)(a4+a2b2+b4)=(a+b)(a−b)(a4+a2b2+b4)

Factorise:

$a^{6} - b^{6}$

$W e k n o w a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) a^{6} - b^{6} = (a^{2})^{3} - (b^{2})^{3} = (a^{2} - b^{2}) ((a^{2})^{2} + (a^{2}) \times (b^{2}) + (b^{2})^{2}) = (a^{2} - b^{2}) (a^{4} + a^{2} b^{2} + b^{4}) = (a + b) (a - b) (a^{4} + a^{2} b^{2} + b^{4})$