Let \(x^{\frac{1}{3}} = b\) so that the equation is
\((x^{\frac{1}{3}})^{2} - 5x^{\frac{1}{3}} + 6 = 0\)
= \(b^{2} - 5b + 6 = 0\)
\(b^{2} - 3b - 2b + 6 = 0\)
\(b(b - 3) - 2(b - 3) = 0 \implies b = \text{2 or 3}\)
\(x^{\frac{1}{3}} = 2 \implies x = 2^{3} = 8\)
\(x^{\frac{1}{3}} = 3 \implies x = 3^{3} = 27\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}