factorize

i suggest you give us the original problem. This looks like a problem that needs a substitution, not factorization

Using Wolframalpha

bhatt sir is saying Right. you have to put original question.

may be which is in Trig. form

thanks

z³ = 18 + 26i , where z = x + iy and x,y are integers. Then find x and y.

\hspace{-16}$Here $\mathbf{Z^3=18+26i}$\\\\ Put $\mathbf{z=x+iy}$\\\\ $\mathbf{(x+iy)^3=18+26i}$\\\\ $\mathbf{x^3-iy^3+3ix^2y-3xy^2=18+26i}$\\\\ $\mathbf{(x^3-3xy^2)+i(3x^2y-y^3)=18+26i}$\\\\ Camparaing Real and Imaginary parts\\\\ $\mathbf{x^3-3xy^2=18............................(1)}$\\\\ $\mathbf{3xy^2-y^3=26............................(2)}$\\\\ $\mathbf{x=r.\cos \theta}$ and $\mathbf{y=r.\sin \theta}$\\\\ $\mathbf{r^3.(\cos^3 \theta-3.\cos \theta.\sin^2 \theta)=18}$\\\\ $\mathbf{r^3.\cos 3\theta =18.......................(3)}$\\\\ Similarly \\\\ $\mathbf{r^3.\sin 3\theta=26........................(4)}$\\\\ $\tan 3\theta =\frac{26}{18}=\frac{13}{9}$\\\\ and $\mathbf{\sin 3\theta =\frac{13}{5\sqrt{10}}}$ and $\mathbf{\cos 3\theta =\frac{9}{5\sqrt{10}}}$\\\\ So $\mathbf{r^3=10\sqrt{10}=(\sqrt{10})^3}$\\\\ So $\mathbf{r=\sqrt{10}}$\\\\ So $\mathbf{\cos 3\theta =\frac{18}{10.\sqrt{10}}}$\\\\ Now How can We solve for $\mathbf{\theta}$

@man111 ,
1)just a typo in ur eqn (2) it should be 3x²y.
2) i think unless it is mentioned that x²+y²=r² ,how can we restrict x,y by substituting x=rcos\theta and y=rsin\theta?