Question

$(216{)}^{\frac{3}{5}}\times (2500{)}^{\frac{2}{5}}\times (300{)}^{\frac{1}{5}}$ as a product of its primes is

• A. $2^3\times 3^7\times 5^3$
• B. $2^3\times 3^2\times 5^2$
• C. $2^4\times 3^2\times 5^2$
• D. $2^5\times 3^4\times 5^2$

Answer 1

The correct option is B

$2^3\times 3^2\times 5^2$

$(216{)}^{\frac{3}{5}}\times (2500{)}^{\frac{2}{5}}\times (300{)}^{\frac{1}{5}}$

$=(2^3\times 3^3{)}^{\frac{3}{5}}\times (2^2\times 5^4{)}^{\frac{2}{5}}\times (2^2\times 3\times 5^2{)}^{\frac{1}{5}}$

$=(2^3{)}^{\frac{3}{5}}\times (3^3{)}^{\frac{3}{5}}\times (2^2{)}^{\frac{2}{5}}\times (5^4{)}^{\frac{2}{5}}\times (2^2{)}^{\frac{1}{5}}\times (3{)}^{\frac{1}{5}}\times (5^2{)}^{\frac{1}{5}}$
$[\because (ab{)}^m=\left(a^m\right)\times \left(b^m\right)]$

$=2^{\frac{9}{5}}\times 3^{\frac{9}{5}}\times 2^{\frac{4}{5}}\times 5^{\frac{8}{5}}\times 2^{\frac{2}{5}}\times 3^{\frac{1}{5}}\times 5^{\frac{2}{5}}$
$[\because (a^m{)}^n=\left(a{)}^{mn}\right]$

$=2^{(\frac{9}{5}+\frac{4}{5}+\frac{2}{5})}\times 3^{(\frac{9}{5}+\frac{1}{5})}\times 5^{(\frac{8}{5}+\frac{2}{5})}$
$[\because a^m\times a^n=a^{(m+n)}]$

$=2^{\frac{15}{5}}\times 3^{\frac{10}{5}}\times 5^{\frac{10}{5}}$

$=2^3\times 3^2\times 5^2$