Question

if (4a + 3b) = 10 and ab = 2 , find the value of (64 a³ + 27 b³)​

Answer 1

Given -

$(4a + 3b) = 10 \: and \: ab = 2$

To find -

${64a}^{3} + {27b}^{3}$

Solution -

$(4a + 3b) = 10$

$Cubing \: \: both \: \: sides$

${(4a + 3b)}^{3} = {10}^{3}$

$By \: applying \: the \: formula \: {(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b) \: we \: will \: get$

${(4a)}^{3} + {(3b)}^{3} + 3 \times 4a \times 3b(4a + 3b) = 1000$

${64a}^{3} + {27b}^{3} + 36ab(4a + 3b) = 1000$

${64a}^{3} + {27b}^{3} + 36 \times 2 \times 10 = 1000$

${64a}^{3} + {27b}^{3} + 720 = 1000$

${64a}^{3} + {27b}^{3} = 1000 - 720$

${64a}^{3} + {27b}^{3 } =280$

$Therefore \: the \: value \: of \: {64a}^{3} + {27b}^{3} \: is \: 280$