Question

If x is equal to 3 + 3 raised to the power 1 by 3 + 3 raise to the power 2 by 3 then show that x cube minus 9 X square + 18 x minus 12 is equal to zero

Answer 1

$\textbf{Given:}$

$x=3+3^{\frac{1}{3}}+3^{\frac{2}{3}}$

$\textbf{To show:}$

$x^3-9x^2+18x-12=0$

$\textbf{Solution:}$

$\text{Consider,}$

$x=3+3^{\frac{1}{3}}+3^{\frac{2}{3}}$

$\implies\,x-3=3^{\frac{1}{3}}+3^{\frac{2}{3}}$......(1)

$\text{Cubing on bothsides, we get}$

$\implies(x-3)^3=(3^{\frac{1}{3}}+3^{\frac{2}{3}})^3$

$\text{Using,}$

$\boxed{\bf\,(a+b)^3=a^3+b^3+3ab(a+b)}$

$\boxed{\bf\,(a-b)^3=a^3-b^3-3ab(a-b)}$ $\text{we get}$

$\implies\,x^3-27-3(3)(x)(x-3)=3+3^2+3(3^{\frac{1}{3}}.3^{\frac{2}{3}})(3^{\frac{1}{3}}+3^{\frac{2}{3}})$

$\text{Using (1), we get}$

$\implies\,x^3-27-9x(x-3)=3+9+3(3)(x-3)$

$\implies\,x^3-27-9x^2+27x=12+9(x-3)$

$\implies\,x^3-27-9x^2+27x=12+9x-27$

$\implies\,x^3-9x^2+27x=12+9x$

$\implies\bf\,x^3-9x^2+18x-12=0$

$\textbf{Hence proved}$

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Answer 2

Answer:

0 is answer

exact answer