Question

$\sqrt{2 {x}^{3} } + \sqrt[3]{2}{x}^{2} + 6x \div 2x$
solve this answer please

Answer 1

Hello:
√2x³+∛2x²+6x÷2x;
(√2x³)²+(∛2x²)³+(6x÷2x);
2x³+2x²+3x;
x(2x²+2x+3)

bye :-)

Answer 2

The question is not very clear. Perhaps 2x divides the entire expression, all terms.  So we need to simplify the terms.

$\frac{\sqrt{2x^2}+\sqrt[3]{2x^2}+6x}{2x}\\\\=\frac{(2x^2)^\frac{1}{2}+(2x^2)^\frac{1}{3}+6x}{2x}\\\\=2^{\frac{1}{2}-1} \: x^{2*\frac{1}{2}-1}+2^{\frac{1}{3}-1} * x^{2*\frac{1}{3}-1}+3 \\\\=\sqrt{2}*x^0+\frac{1}{\sqrt[3]{2^2 \: x}}+3$