Question

If $a=\frac{4}{3}$ and $b=\frac{3}{4}$, then the sum of the odd terms in $(6a+8b{)}^3$ exceeds the sum of the even terms by

• A. 1
• B. 8
• C. 27
• D. 81

Answer 1

The correct option is B 8

$(6a+8b{)}^3$
$=8(3a+4b{)}^3$
$=8(27a^3+36ab(3a+4b)+64b^3)$
$=8(27a^3+108a^{2}b+144a{b}^2+64b^3)$
Therefore sum of odd terms $-$ sum of even terms
$=8\left(\right(27a^3+144a{b}^2)-(108a^{2}b+64b^3\left)\right)$
Substituting $b=\frac{3}{4}$ and $a=\frac{4}{3}$
$=8(64+108-(144+27\left)\right)$
$=8(64-63)$
$=8$