Solve sqrt{-2^2}=left|-1right|+1/9*-3^2+sqrt[3]{-8}

$$\sqrt{ { -2 }^{ 2 } } = \left| -1 \right| + \frac{ 1 }{ 9 } \times { -3 }^{ 2 } + \sqrt[ 3 ]{ -8 }$$

Verify

Solution Steps

Calculate -2 to the power of 2 and get 4.

$$\sqrt{4}=|-1|+\frac{1}{9}\left(-3\right)^{2}+\sqrt[3]{-8}$$

Calculate the square root of 4 and get 2.

$$2=|-1|+\frac{1}{9}\left(-3\right)^{2}+\sqrt[3]{-8}$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -1 is 1.

$$2=1+\frac{1}{9}\left(-3\right)^{2}+\sqrt[3]{-8}$$

Calculate -3 to the power of 2 and get 9.

$$2=1+\frac{1}{9}\times 9+\sqrt[3]{-8}$$

Multiply \frac{1}{9} and 9 to get 1.

$$2=1+1+\sqrt[3]{-8}$$

Add 1 and 1 to get 2.

$$2=2+\sqrt[3]{-8}$$

Calculate \sqrt[3]{-8} and get -2.

$$2=2-2$$

Subtract 2 from 2 to get 0.

$$2=0$$

Compare 2 and 0.

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