Proof using properties of an isosceles or right-angle triangle

Question

Given a $\triangle ABC$ with sides $AB=BC$ and $\angle B=100^\circ $, prove that $$a^3 + b^3 = 3a^2b$$ where $a=AB=BC$ and $b=AC$,

I have tried to use simultaneously the sine and cosine rules as well as the Pythagorean Theorem with all my attempts failing to prove that $LHS =RHS$. I would greatly appreciate a hint on how to prove the proposition.

Answer 1

A straight forward application of cosine rule should tell you that $$ b = 2a\sin(50) $$

Consider

$$ \begin{equation} \begin{split} a^3 + b^3 - 3a^2b & = a^3(1+8\sin^350-6\sin50) \\ & = a^3(1+8\frac{(3\sin50 - \sin 30)}{4}-6\sin50) \\ & = a^3(1+6\sin50-2\sin30-6\sin50) \\ & = 0 \end{split} \end{equation} $$

Answer 2

$a^3+b^3=3a^2b$ using cosine

$b^2=a^2+a^2-2aa\cos \beta,\beta<100$

$b^2=2a^2(1-\cos \beta)\\let \\(\frac{b}{a})^2=t^2=2(1-\cos \beta)$

$a^3+b^3=3a^2b\implies 1+t^3=3t\implies t^3-3t+1=0$

$t(t^2-3)=-1\implies t^2(t^2-3)^2=2(1-\cos \beta)(1-\cos\beta-3)^2\\=2(1-\cos \beta)(1+4\cos \beta+4\cos^2 \beta)\\=2(1+3\cos\beta-4\cos^3\beta)=2+6\cos\beta-8\cos^3\beta=1$

now we should show that

$8\cos^3\beta-6\cos\beta-1=2\cos3\beta-1=0 $

$3\beta=\pm \frac{\pi}{3}+2\pi k,k\in Z$