If $ab(a+b)=2$ then what will be the value of $\frac{1}{(ab)^3}-a^3-b^3$?

Question

The question given is,

If $ab(a+b)=2$ then show that $\dfrac{1}{(ab)^3}-a^3-b^3=?$
$(a)\ \ \ \ \ 0$
$(b)\ \ \ \ \ 1$
$(c)\ \ \ \ \ 3$
$(d)-1$

I noted that $a=1$ and $b=1$ satisfy the given condition. Then I tried to replace $a^3+b^3$ with $(a+b)^3-3ab(a+b)=(a+b)^3-6$, and then replaced $(a+b)$ with $\dfrac{2}{ab}$ but I couldn't eliminate $a$ and $b$.

Any hints will be appreciated, thanks.

Edit: I can't recall the exact four options but the question statement is same.

Answer 1

I haven't gone through all the details, but let me give you a hint.

$$ ab(a+b) = 2 \Rightarrow \frac{1}{ab} = \frac{1}{2}(a+b) $$

$$ \Rightarrow \frac{1}{(ab)^3} - a^3 - b^3 = \frac{1}{8}(a+b)^3 - (a^3 + b^3). $$

This is similar to what you had suggested, but you will probably find it easier working with $a^r$ and $b^r$ rather than their inverses.

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Actually, the question is incorrect. I tested it on multiple pairs on Matlab, and it fails... This can be seen as for large $a$ the right-hand side is like $-(7/8)a^3 \rightarrow -\infty$. (Note that $a \rightarrow -\infty \Rightarrow b \rightarrow 0$ so that the other equality holds, but this doesn't matter as in the final expression, the $a$ is not in a product with $b$.)

Basically, my advice is to look for the correct version of the question, then use the advice that I gave above the line.

Hope this helps! :)

Answer 2

If $ab(a+b)=2$ then we have the identity $ab=\frac{2}{a+b}$. So $$\dfrac{1}{(ab)^3} = \dfrac{(a+b)^3}{8} = \frac18(a^3 + b^3 + 3ab(b + a))$$

Substituting again for $ab$ we get: $\frac18(a^3 + b^3 + 3\frac{2}{a+b} (b + a)) = \frac18(a^3 + b^3 + 6)$, which is in general not equal to $a^3 + b^3 - 1$, so the question is impossible to answer.

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EDIT: $\frac18(a^3 + b^3 + 6)$ is in general not equal to $a^3 + b^3 + c$ for any constant $c$. This is because the second polynomial grows faster. That in turn is because the coefficients of the cubic terms are bigger in the second term. So the question is still impossible to answer.

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EDIT: Note, the original identity is only possible because we can prove that $a+b \neq 0$. Without this, we wouldn't be allowed to take a $0$ term to the denominator of the fraction in the identity. So why can we assert that it's OK in this case? Well, if the sum $a+b$ was $0$, then $ab(a+b) = ab\cdot0 = 0 \neq 2$. This violated the original identity we are given, meaning the sum $a+b$ can't be zero, and we're OK to do that algebraic manipulation of dividing by the $a+b$.

Answer 3

I will interpret the question in the following sense: given the constraint $ab(a+b)=2$, which of the values on the list can $\frac{1}{(ab)^3}-a^3-b^3$ take?

Due to the given constraint, any possible value $x$ of the expression satisfies

$x = \frac{1-a^3b^6-a^6b^3}{a^3b^3} = \frac{1-a^3b^3(a+b)^3+3a^4b^4(a+b)}{a^3b^3} = \frac{1-2^3+6a^3b^3}{a^3b^3} = 6-\frac{7}{a^3b^3}$.

For any particular value of $x$, this means that $ab=\sqrt[3]{\frac{7}{6-x}}$ and therefore $a+b = \frac{2}{ab} = 2\sqrt[3]{\frac{6-x}{7}}$. From this we get that $a$ and $b$ must be the solutions for $z$ of the quadratic equation

$z^2-2\sqrt[3]{\frac{6-x}{7}}z+\sqrt[3]{\frac{7}{6-x}} = 0$, whose discriminant is

$\sqrt{\sqrt[3]{\frac{6-x}{7}}^2 - \sqrt[3]{\frac{7}{6-x}}} = \sqrt{\sqrt[3]{\frac{7}{6-x}}(\frac{6-x}{7}-1)} = \sqrt{\sqrt[3]{\frac{7}{6-x}}\frac{-1-x}{7}}$,

which is not a real number for $-1<x<6$. So from the given choices, only $-1$ can occur, in which case we get $ab=1,a+b=2\ \Rightarrow\ a=b=1$.