Evaluate the values of $x$ in $\sqrt{2x-5} = \sqrt[3]{6x-15}$ 
$$\sqrt{2x-5} = \sqrt[3]{6x-15}$$
  
  
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*Evaluate the values of $x$

I believe that there would be an easier approach to this problem because I will have to expand 3th degree binomial as shown below
$$(2x-5)^3 = (6x-15)^2$$
What am I missing?
Regards
 A: $$(2x-5)^3 = (6x-15)^2$$
$$(2x-5)^3 = (3(2x-5))^2$$
$$(2x-5)^3 = 9(2x-5)^2$$
This may help :)
Then: $(2x-5)=0\lor (2x-5)=9$
Thus: $x=2\frac{1}{2}\lor x=7$
A: Hint: It is $$(2x-5)^3=9(2x-5)^2$$ and $$(2x-5)^2(2x-14)=0$$
since $$(2x-5)^3-9(2x-5)^2=(2x-5)^2(2x-5-9)=0$$
A: 
because I will have to expand 3th degree binomial as shown below: $(2x-5)^3 = (6x-15)^2$

Which is not a difficult thing to do:
$8x^3 - 60x^2 + 150x - 125 = 36x^2 - 180x +225$ so 
$8x^3 -96x^2 + 330x - 350=0$
which is not an unreasonable thing to expect a student to be able to factor and solve (although I pity the student who tries).
However the student should at this point see:
$(2x -5)^3 = (6x-15)^2 = (3(2x-5))^2 = 9(2x-5)^2$ 
So either $2x-5 =0$ or $2x -5 = 9$.
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.... or one can do it the sweaty he-man way:
$\sqrt{2x - 5} = \sqrt[3]{6x - 15} = \sqrt[3]3\sqrt[3]{2x - 5}$.
If $x - 5 = 0$ we have $x = 2 \frac 12$ as a potential answer.
If we assume $2x -5\ne 0$ we have
$\frac {\sqrt{2x - 5}}{\sqrt[3]{2x - 5}} = \sqrt[3] 3$.
Assume $2x - 5 > 0$ (which we must for $2x - 5 \ne 0$ and $\sqrt{2x - 5}$ to be defined) we have:
$\frac {(2x-5)^{\frac 12}}{(2x-5)^{\frac 13}}=(2x-5)^{\frac 12 - \frac 13} = (2x-5)^{\frac 16} = 3^{\frac 13}$
So $2x-5 = (3^{\frac 13})^6 = 3^2 = 9$.
So $x = 7$.
So solutions are
$x = 2\frac 12, 7$.
