Solve sqrt{116/9}+sqrt[3]{-343}+sqrt[n]{7^2n}=

$$\sqrt { 1 \frac { 16 } { 9 } } + \sqrt[ 3 ] { - 343 } + \sqrt[ n ] { 7 ^ { 2 n } } =$$

Factor

$$\frac{131}{3} = 43\frac{2}{3} = 43.666666666666664$$

Solution Steps

Multiply 1 and 9 to get 9.

$$factor(\sqrt{\frac{9+16}{9}}+\sqrt[3]{-343}+\sqrt[n]{7^{2n}})$$

Add 9 and 16 to get 25.

$$factor(\sqrt{\frac{25}{9}}+\sqrt[3]{-343}+\sqrt[n]{7^{2n}})$$

Rewrite the square root of the division \frac{25}{9} as the division of square roots \frac{\sqrt{25}}{\sqrt{9}}. Take the square root of both numerator and denominator.

$$factor(\frac{5}{3}+\sqrt[3]{-343}+\sqrt[n]{7^{2n}})$$

Calculate \sqrt[3]{-343} and get -7.

$$factor(\frac{5}{3}-7+\sqrt[n]{7^{2n}})$$

Subtract 7 from \frac{5}{3} to get -\frac{16}{3}.

$$factor(-\frac{16}{3}+\sqrt[n]{7^{2n}})$$

Factor out \frac{1}{3}.

$$\frac{-16+3\times \left(7^{2n}\right)^{\frac{1}{n}}}{3}$$

Evaluate

$$\frac{131}{3}\approx 43.666666667$$