Question

If a, b and c are positive real numbers such that $a^3+\frac{b^3}{8}+\frac{c^3}{27}=\frac{1}{2}abc$ then a : b : c is equal to

• A. 1 : 8 : 27
• B. 1 : 2 : 3
• C. 1 : 4 : 9
• D. 3 : 2 : 1

Answer 1

The correct option is B 1 : 2 : 3

$a^3+(\frac{b}{2}{)}^3+(\frac{c}{3}{)}^3=\frac{abc}{2}$
$\Rightarrow a^3+(\frac{b}{2}{)}^3+(\frac{c}{3}{)}^3=3a\left(\frac{b}{2}\right)\left(\frac{c}{3}\right)$
Use the identity $x^3+y^3+z^3–3xyz=(x+y+z)(x^2+y^2+z^2–xy–yz–zx).$
If $x+y+z=3xyz$, then either $x+y+z=0$ or $x^2+y^2+z^2–xy–yz–zx=0$.
$\therefore a+\frac{b}{2}+\frac{c}{3}=0$ or $a^2+(\frac{b}{2}{)}^2+(\frac{c}{3}{)}^2-\frac{ab}{2}-\frac{bc}{6}-\frac{ac}{3}=0$
Since, a, b, c are positive real numbers, the sum $a+\frac{b}{2}+\frac{c}{3}$ cannot be zero.
$\therefore a^2+\frac{b^2}{4}+\frac{c^2}{9}-\frac{ab}{2}-\frac{bc}{6}-\frac{ac}{3}=0$
$\Rightarrow \frac{1}{2}\left[2\right(a^2+\frac{b^2}{4}+\frac{c^2}{9}-\frac{ab}{2}-\frac{bc}{6}-\frac{ac}{3}\left)\right]=0$
$\Rightarrow \frac{1}{2}[a^2+\frac{b^2}{4}-ab+\frac{b^2}{4}+\frac{c^2}{9}-\frac{bc}{3}+a^2+\frac{c^2}{9}-\frac{2ac}{3})]=0$
$\Rightarrow \frac{1}{2}[a^2+(\frac{b}{2}{)}^2-2\times a\times \frac{b}{2}+(\frac{b}{2}{)}^2+(\frac{c}{3}{)}^2]$
$\Rightarrow \frac{1}{2}\left[\right(a-\frac{b}{2}{)}^2+(\frac{b}{2}-\frac{c}{3})+(a-\frac{c}{3}{)}^2]=0$
$\therefore a-\frac{b}{2}=0,\frac{b}{2}-\frac{c}{3}=0,a-\frac{c}{3}=0$
$\Rightarrow a=\frac{b}{2},\frac{b}{2}=\frac{c}{3},a=\frac{c}{3}$
$\Rightarrow a=\frac{b}{2}=\frac{c}{3}$
Let $a=\frac{b}{2}=\frac{c}{3}=k$
Then, a = k, b = 2k and c = 3k.
$\therefore a:b:c=k:2k:3k$
= 1 : 2 : 3
Hence, the correct answer is option (2).