Question

divide 64a^3+27b^3 by (4a+3b)​

Answer 1

Answer :-

$16a^{2} + 12ab + 9b^{2}$

Solution :-

$\frac{64a^{3} + 27b^{3}}{4a + 3b}$

$\frac{(4a)^{3} + (3b)^{3} }{4a + 3b}$

Expand: $a^{3} + b^{3} = (a+b) (a^{2} + ab + b^{2})$

$\frac{(4a + 3b) (16a^{2} + 12ab + 9b^{2})}{4a + 3b}$

$16a^{2} + 12ab + 9b^{2}$

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