# Cubed exponent equation

$$\left(2 · 3^x\right)^3 + \left(9^x − 3\right)^3 = \left(9^x + 2 · 3^x − 3\right)^3$$ Solve the equation

I got the answer to the problem, in which I evaluated the whole expression (which was quite hard) which was $0$ and $1/2$, is there a way to solve this problem in a less tedious way?

• I do not understand what you mean by "I evaluated the whole expression". How exactly did you solve this for $x$? Commented Oct 26, 2017 at 12:57

Note that the entire equation is of the form $$a^3 + b^3 = (a+b)^3$$ with $a = 2\cdot 3^x$ and $b = 9^x - 3$. This reduces to $$0 = 3a^2b + 3ab^2\\ 0 = 3ab(a+b)$$ which has the three solutions $a = 0$, $b = 0$, or $a+b = 0$. Clearly, $a = 0$ can't happen. $b = 0$ means $9^x = 3$, which has the solution $x = \frac12$. Finally, $a+b = 0$ means $$2\cdot 3^x + 9^x - 3 = 0\\ (3^x)^2 + 2\cdot 3^x - 3 = 0\\ 3^x = \frac{-2\pm\sqrt{2^2 - 4\cdot 1\cdot (-3)}}{2} = -1\pm 2$$ Since $3^x$ can't be negative, we can only keep $3^x = 1$, which gives $x = 0$.

• Is it valid to distribute the exponent like that in your first equation? I mean I don't remember that in algebra. Commented Oct 26, 2017 at 23:35
• @MARXOS Not in general. But that is exactly what OP's equation says. In other words, it's only valid for very specific values of $a$ and $b$. Not a property of algebra. Commented Oct 26, 2017 at 23:37
• @MARXOS I could've inserted the line $$a^3+b^3=a^3+3a^2b+3b^2a+b^3$$ between $$a^3+b^3=(a+b)^3$$ and $$0=3a^2b+3ab^2$$ Is that what you were asking about? Or were you asking about the construction of $a^3+b^3=(a+b)^3$ itself? Commented Oct 27, 2017 at 5:42
• @Arthur: Was asking about the construct of trying to distribute the exponent into a parenthetical expression. Commented Oct 27, 2017 at 5:49

First of all, take $z=3^x$. That way, you obtain the equation: $$(2z)^3 + (z^2 − 3)^3 = (z^2 + 2z − 3)^3.$$ Then try to factor it using your favourite method (Rufini, for instance), so that you obtain: $$z (z - 1) (z + 3) (z^2 - 3)=0.$$ Solve for $z$ and use those to compute $x$ knowing that $z=3^x$.

Hint. The left side is the sum of two cubes, and $$a^3 + b^3 = (a+b)(a^2 -ab + b^2) .$$

• The approach by Arthur is more efficient.
– user65203
Commented Oct 26, 2017 at 13:07

Using $t:=3^x$ and following the factorization by @Arthur ($(a+b)^3-a^3-b^3=3ab(a+b)$), the equation is equivalent to

$$t(t^2-3)(t^2+2t-3)=t(t-\sqrt 3)(t+\sqrt 3)(t-1)(t+3)=0.$$

Only the positive $t$ yield a solution and we immediately have

$$x=\frac12\text{ or }x=0.$$

HINT: You can use the identity $$x^3+y^3=(x+y)(x^2-xy+y^2)$$ on the left hand side.

Preferably after you let $3^x=a$.

• thank you. I just realised the sign :) Commented Oct 26, 2017 at 13:05

Some hints:

Let $u=3^x$. Then the equation becomes $(2u)^3+(u^2-3)^3=(u^2+2u-3)^3$. You can rearrange this and it factorises as $6u(u-1)(u+3)(u^2-3)=0$.