Question

Factorise 8a cube - b cube -12a squre b +6ab square

Answer 1

To factorise:

$8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}$

$=(2 a)^{3}+(-b)^{3}+3 .(2 a)^{2} \cdot(-b)+3.2 a \cdot(-b)^{2}$

$=(2 a-b)^{3}$

               $\left[\text { Since },(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}\right]$

Factorization is defined as process of finding the least factor or least divisible factor by which the given number is divisible. They can be either a prime factor or any whole number which divides the number exactly. Factorization is of two types prime factorization and normal L-division method.

Answer 2

$\bold{\bf{\underline{\underline{Answer:}}}}$

$\bold {\bf {∴\:(2a + b) (2a + b) (2a + b)} \:are\:the\:factors.}$

$\bold{\underline {Given:}}$

$⟹ {8a}^{3} - {b}^{3} - {12a}^{2}b + 6a{b}^{2}$

$\bold{\underline {To\:Find:}}$

${⟹}\:Factorise\: the \:given\: equation=\:?$

$\bold{\bf{\underline{\underline{Step \:by\: step \:explaination :}}}}$

$⟹ {{8a}^{3}} - {{b} ^{3}} - {{12a}^{2}b} + {6a{b}^{2}}$

$\bf{By, {(a + b)}^{3}= {a} ^{3} + {3a}^{2}b + 3a{b}^{2} + {b}^{3}}$

$\bf{Now,}$

$⟹ {{8a}^{3}} - {{b} ^{3}} - {{12a}^{2}}b + {6a{b}^{2}}$

$⟹ {{2a}^{3}} - {{b}^{3}} + 3 × {{2a}^{3}} b+ 3 × 2a{{b}^{3}}$

$\bf{According \:to\: above \:identity\:⟹}$

$⟹ {(2a + b)}^{3}$

$⟹ {(2a + b)}^{3} \:{(2a + b)}^{3} \:{(2a + b)}^{3}$

$\bf{Hence, {(2a + b)}^{3}\:{(2a + b)}^{3}\:{(2a + b)}^{3} are\: the\: factors.}$

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