Question

Factorize $64a^3–343b^3$

• A. $(4a–7b)(16a^2+49b^2+28ab)$
• B. $(4a+7b)(16a^2-49b^2+28ab)$
• C. $(4a–7b)(16a^2+49b^2-28ab)$
• D. $(4a+7b)(16a^2+49b^2+28ab)$

Answer 1

The correct option is A

$(4a–7b)(16a^2+49b^2+28ab)$

$64a^3–343b^3={\left(4a\right)}^3–{\left(7b\right)}^3$

We know that,
$x^3–y^3$ =$(x–y$)$(x^2+xy+y^2$)

$\therefore {\left(4a\right)}^3–{\left(7b\right)}^3$
$=(4a–7b)[{\left(4a\right)}^2+(7b\left)\right(4a)+{\left(7b\right)}^2]$
$=(4a–7b)(16a^2+49b^2+28ab)$