Question

Is 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 always equal to

(1) 1 : 8 : 27
(2) 1 : 2 : 3
(3) 1 : 4 : 9
(4) 3 : 2 : 1

Answer 1

It's a pretty long sum

$a^3+\frac{b^3}{8}+\frac{c^3}{27}-\frac{1}{2}abc=0 \\3*\frac{b}{2}*\frac{c}{3}=\frac{1}{2}abc$

So by cubic identity we can infer that,

$(a+\frac{b}{2}+\frac{c}{3})(a^2+\frac{b^2}{2}+\frac{c^2}{9}-\frac{ab}{2}-\frac{bc}{6}-\frac{ac}{3})=0$

But, $(a+\frac{b}{2}+\frac{c}{3})$ cannot be 0 because a, b and c are positive

So $(a^2+\frac{b^2}{2}+\frac{c^2}{9}-\frac{ab}{2}-\frac{bc}{6}-\frac{ac}{3})$ has to be zero

Doubling both sides and using $(a+b)^2$ identity we will get,

$(a-\frac{b}{2})^2+(\frac{b}{2}-\frac{c}{3})^2+(\frac{c}{3}-a)^2=0$

As sum of two or more squares cannot be zero, each of these terms are zero,

So we get $a-\frac{b}{2}=0$, $\frac{b}{2}-\frac{c}{3}=0$ and $\frac{c}{3}-a=0$

Now when we solve each of these linear equations, we will get,

$2a=b$

$3b=2c$ (we don't need to use this one tho)

$3a=c$

Now when we make the ratio between these three we will get,

$a:2a:3a$

which simplifies to 1 : 2 : 3

Therefore the answer is (2) 1 : 2 : 3

HOPE IT HELPS!

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