Solve left(2^3-2^-3right)^2=2^6*2({2}^3*{2}^-3)+{left({2}^-3right)}^2

$${ \left( { 2 }^{ 3 } - { 2 }^{ -3 } \right) }^{ 2 } = { 2 }^{ 6 } \times 2( { 2 }^{ 3 } \times { 2 }^{ -3 } )+ { \left( { 2 }^{ -3 } \right) }^{ 2 }$$

Verify

Solution Steps

To multiply powers of the same base, add their exponents. Add 6 and 1 to get 7.

$$\left(2^{3}-2^{-3}\right)^{2}=2^{7}\times 2^{3}\times 2^{-3}+\left(2^{-3}\right)^{2}$$

To multiply powers of the same base, add their exponents. Add 7 and 3 to get 10.

$$\left(2^{3}-2^{-3}\right)^{2}=2^{10}\times 2^{-3}+\left(2^{-3}\right)^{2}$$

To multiply powers of the same base, add their exponents. Add 10 and -3 to get 7.

$$\left(2^{3}-2^{-3}\right)^{2}=2^{7}+\left(2^{-3}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply -3 and 2 to get -6.

$$\left(2^{3}-2^{-3}\right)^{2}=2^{7}+2^{-6}$$

Calculate 2 to the power of 3 and get 8.

$$\left(8-2^{-3}\right)^{2}=2^{7}+2^{-6}$$

Calculate 2 to the power of -3 and get \frac{1}{8}.

$$\left(8-\frac{1}{8}\right)^{2}=2^{7}+2^{-6}$$

Subtract \frac{1}{8} from 8 to get \frac{63}{8}.

$$\left(\frac{63}{8}\right)^{2}=2^{7}+2^{-6}$$

Calculate \frac{63}{8} to the power of 2 and get \frac{3969}{64}.

$$\frac{3969}{64}=2^{7}+2^{-6}$$

Calculate 2 to the power of 7 and get 128.

$$\frac{3969}{64}=128+2^{-6}$$

Calculate 2 to the power of -6 and get \frac{1}{64}.

$$\frac{3969}{64}=128+\frac{1}{64}$$

Add 128 and \frac{1}{64} to get \frac{8193}{64}.

$$\frac{3969}{64}=\frac{8193}{64}$$

Compare \frac{3969}{64} and \frac{8193}{64}.

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