CoolStuffYT

As EP said, substitute \(x=3\) and \(x=-3\) into your equations, and solve from there.

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Multiply by \(\frac{x+1}{x+1}\) and get \(\frac{5(x+1)}{\frac{x+1}{6}+1}\), then multiply by \(\frac{6}{6}\) and get \(\frac{30x+6}{x+7}\) .

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\(f(-2)=(-2)^3-2(-2)=-8+4=-4\)

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See here:

https://web2.0calc.com/questions/please-help-10_2#r2

These questions are from AoPS, a math site. I am doing it, and these are the homework problems. AoPS specifically said not to post any of these questions on the site, so just saying.

Oh yes, this is from the Geometry Homework on AoPS.

I can't tell you the answer but...

Are there any triangles you can use?

Smart triangles? like 30-60-90?

Is it a coincidence that \(AB=\sqrt3\)? Maybe it's part of a 30-60-90 triangle.

You should be able to find \(DR\) with these hints.

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We see a difference of squares! \(\sqrt9=3, \sqrt4=2\) so...

\((3x+2)(3x-2)\).

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You can see that it is \((x+6)(x+7)\)

or

Quadratic Formula

\(\frac{-13\pm \sqrt{13^2-4\times 42}}{2}=\frac{-13+1}{2}=-6, \frac{-13-1}{2}=-7\) so it's \((x+6)(x+7)\) 

(wow, for me, that was a lot of complicated LaTeX!)

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Hopefully you know that \(1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2\).

Therefore, we see that \(n=7\) when the equations are equal so \(n=6\)

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Sorry, I'm not sure.

You could search up how to make the cube root command.

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