Question

if k = 13 power 2/3 +13 power 1/3 + 1​

Answer 1

$\textbf{Given:}$

$\mathsf{k=13^\frac{2}{3}+13^\frac{1}{3}+1}$

$\textbf{To prove:}$

$\mathsf{k^3-3k^2-37k-144=0}$

$\textbf{Solution:}$

$\mathsf{Consider,}$

$\mathsf{k=13^\frac{2}{3}+13^\frac{1}{3}+1}$

$\mathsf{k-1=13^\frac{2}{3}+13^\frac{1}{3}}$

$\textsf{cubing on bothsides, we get}$

$\mathsf{(k-1)^3=(13^\frac{2}{3}+13^\frac{1}{3})^3}$

$\mathsf{(k-1)^3=(13^\frac{2}{3})^3+(13^\frac{1}{3})^3+3(13^\frac{2}{3})(13^\frac{1}{3})(13^\frac{2}{3}+13^\frac{1}{3})}$

$\mathsf{(k-1)^3=(13^\frac{2}{3})^3+(13^\frac{1}{3})^3+3(13)(k-1)}$

$\mathsf{(k-1)^3=13^2+13+39(k-1)}$

$\mathsf{k^3-1^3-3k(k-1)=169+13+39k-39}$

$\mathsf{k^3-1-3k^2+3k=143+39k}$

$\implies\boxed{\mathsf{k^3-3k^2-37k-144=0}}$

$\textbf{Find more:}$

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

https://brainly.in/question/10830460

Answer 2

Step-by-step explanation:

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