Question

Hello friends..

$factorise \: 1)27a ^{3} + 64b ^{3} \: \: \: \: \: 2)49 y^{3} - 1000 \: using \: \: the \: \: identity.$
please answer my question....

Answer 1

hope this helps you
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a-
$= > 27 {a}^{3} + 64 {b}^{3}$

$= > {(3a)}^{3} + {(4b)}^{3}$

$= > (3a + 4b)(9 {a}^{2} - 12ab + 16 {b}^{2} )$
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b-
$= > 49 {y}^{3} - 1000$

$= > {( \sqrt[3]{49} y) }^{3} - {(10)}^{3}$

$= > ( \sqrt[3]{49} y - 10) \\ ({( \sqrt[3]{49} y })^{2} + ( 10\sqrt[3]{49} y) + 100)$
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by identity
${a}^{3} +{b}^{3} = (a+b)({a}^{2} - ab + {b}^{2})$
${a}^{3} - {b}^{3} = (a-b)({a}^{2} + ab + {b}^{2})$

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