Answer:
[tex]x^{\frac{1}{3}}\:\times \left(x^{\frac{1}{2}}\:+\:2x^2\right)=x^{\frac{5}{6}}+2x^{\frac{7}{3}}[/tex]
Step-by-step explanation:
[tex]x^{\frac{1}{3}}\:\times \left(x^{\frac{1}{2}}\:+\:2x^2\right)[/tex]
Let us simplify the expression
[tex]x^{\frac{1}{3}}\:\times \left(x^{\frac{1}{2}}\:+\:2x^2\right)[/tex]
[tex]\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac[/tex]
[tex]a=x^{\frac{1}{3}},\:b=x^{\frac{1}{2}},\:c=2x^2[/tex]
so
[tex]=x^{\frac{1}{3}}x^{\frac{1}{2}}+x^{\frac{1}{3}}\times \:2x^2[/tex]
[tex]=x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}[/tex]
so the expression becomes
[tex]=x^{\frac{5}{6}}+2x^{\frac{7}{3}}[/tex]
Thus,
[tex]x^{\frac{1}{3}}\:\times \left(x^{\frac{1}{2}}\:+\:2x^2\right)=x^{\frac{5}{6}}+2x^{\frac{7}{3}}[/tex]
