Question

factorise the following by using a suitable identity :a3+b3-8c3+6abc

Answer 1

$\textbf{Given:}$

$a^3+b^3-8c^3+6abc$

$\textbf{To find:}$

$\text{Factors of a^3+b^3-8c^3+6abc }$

$\textbf{Solution:}$

$\text{Consider,}$

$a^3+b^3-8c^3+6abc$

$=a^3+b^3+(-2c)^3-3ab(-2c)$

$\text{Using the algebraic identity,}$

$\boxed{\bf\,a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}$ $\text{we get}$

$=(a+b+(-2c))(a^2+b^2+(-2c)^2-ab-b(-2c)-(-2c)a)$

$=(a+b-2c)(a^2+b^2+4c^2-ab+2bc+2ca)$

$\therefore\bf\,a^3+b^3-8c^3+6abc\\\\=(a+b-2c)(a^2+b^2+4c^2-ab+2bc+2ca)$

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Answer 2

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