Question
Write the value of $\sqrt[3]{7}\times\sqrt[3]{49}.$
✓
Answer
We have to find the value of $\sqrt[3]{7}\times\sqrt[3]{49}.$ So,$\sqrt[3]{7}\times\sqrt[3]{49}=\sqrt[3]{7}\times\sqrt[3]{49}$
$=7^{\frac{1}{3}}\times7^{2\times\frac{1}{3}}$
$=7^{\frac{1}{3}}\times7^{\frac{2}{3}}$
By using law rational exponents $\text{a}^{\text{m}}\times\text{a}^\text{n}=\text{a}^{\text{m}+\text{n}}$ we get,$\sqrt[3]{7}\times\sqrt[3]{49}=7^{\frac{1}{3}}\times7^{\frac{2}{3}}$
$=7^{\frac{1}{3}+\frac{2}{3}}$
$=7^\frac{{1+2}}{3}$
$=7^{\frac{3}{3}}=7$
Hence the value of $\sqrt[3]{7}\times\sqrt[3]{49}$ is 7.
$=7^{\frac{1}{3}}\times7^{2\times\frac{1}{3}}$
$=7^{\frac{1}{3}}\times7^{\frac{2}{3}}$
By using law rational exponents $\text{a}^{\text{m}}\times\text{a}^\text{n}=\text{a}^{\text{m}+\text{n}}$ we get,$\sqrt[3]{7}\times\sqrt[3]{49}=7^{\frac{1}{3}}\times7^{\frac{2}{3}}$
$=7^{\frac{1}{3}+\frac{2}{3}}$
$=7^\frac{{1+2}}{3}$
$=7^{\frac{3}{3}}=7$
Hence the value of $\sqrt[3]{7}\times\sqrt[3]{49}$ is 7.
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