Solve -48=left(1+1right)^2(1)left(1-3right)^2

$$-48= { \left(1+1 \right) }^{ 2 } (1) { \left(1-3 \right) }^{ 2 }$$

Verify

Solution Steps

Add 1 and 1 to get 2.

$$-48=2^{2}\times 1\left(1-3\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$-48=4\times 1\left(1-3\right)^{2}$$

Multiply 4 and 1 to get 4.

$$-48=4\left(1-3\right)^{2}$$

Subtract 3 from 1 to get -2.

$$-48=4\left(-2\right)^{2}$$

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

$$-48=4\times 4$$

Multiply 4 and 4 to get 16.

$$-48=16$$

Compare -48 and 16.

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