Solve sqrt[3]{7^2+5^2-47}+sqrt[4]{[2-(25-4):7]:3*2^3}=

$$\sqrt[ 3 ] { 7 ^ { 2 } + 5 ^ { 2 } - 47 } + \sqrt[ 4 ] { [ 2 - ( 25 - 4 ) : 7 ] : 3 \cdot 2 ^ { 3 } } =$$

Evaluate

$$\text{Indeterminate}$$

Solution Steps

Calculate 7 to the power of 2 and get 49.

$$\sqrt[3]{49+5^{2}-47}+\sqrt[4]{\frac{2-\frac{25-4}{7}}{3}\times 2^{3}}$$

Calculate 5 to the power of 2 and get 25.

$$\sqrt[3]{49+25-47}+\sqrt[4]{\frac{2-\frac{25-4}{7}}{3}\times 2^{3}}$$

Add 49 and 25 to get 74.

$$\sqrt[3]{74-47}+\sqrt[4]{\frac{2-\frac{25-4}{7}}{3}\times 2^{3}}$$

Subtract 47 from 74 to get 27.

$$\sqrt[3]{27}+\sqrt[4]{\frac{2-\frac{25-4}{7}}{3}\times 2^{3}}$$

Calculate \sqrt[3]{27} and get 3.

$$3+\sqrt[4]{\frac{2-\frac{25-4}{7}}{3}\times 2^{3}}$$

Subtract 4 from 25 to get 21.

$$3+\sqrt[4]{\frac{2-\frac{21}{7}}{3}\times 2^{3}}$$

Divide 21 by 7 to get 3.

$$3+\sqrt[4]{\frac{2-3}{3}\times 2^{3}}$$

Subtract 3 from 2 to get -1.

$$3+\sqrt[4]{\frac{-1}{3}\times 2^{3}}$$

Fraction \frac{-1}{3} can be rewritten as -\frac{1}{3} by extracting the negative sign.

$$3+\sqrt[4]{-\frac{1}{3}\times 2^{3}}$$

Calculate 2 to the power of 3 and get 8.

$$3+\sqrt[4]{-\frac{1}{3}\times 8}$$

Multiply -\frac{1}{3} and 8 to get -\frac{8}{3}.

$$3+\sqrt[4]{-\frac{8}{3}}$$