Solve (2sqrt{8}-3sqrt{3}+5sqrt{32})-(3sqrt{27}-6sqrt{24})

$$( 2 \sqrt { 8 } - 3 \sqrt { 3 } + 5 \sqrt { 32 } ) - ( 3 \sqrt { 27 } - 6 \sqrt { 24 } )$$

Evaluate

$$12\left(\sqrt{6}+2\sqrt{2}-\sqrt{3}\right)\approx 42.55039272$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$2\times 2\sqrt{2}-3\sqrt{3}+5\sqrt{32}-\left(3\sqrt{27}-6\sqrt{24}\right)$$

Multiply 2 and 2 to get 4.

$$4\sqrt{2}-3\sqrt{3}+5\sqrt{32}-\left(3\sqrt{27}-6\sqrt{24}\right)$$

Factor 32=4^{2}\times 2. Rewrite the square root of the product \sqrt{4^{2}\times 2} as the product of square roots \sqrt{4^{2}}\sqrt{2}. Take the square root of 4^{2}.

$$4\sqrt{2}-3\sqrt{3}+5\times 4\sqrt{2}-\left(3\sqrt{27}-6\sqrt{24}\right)$$

Multiply 5 and 4 to get 20.

$$4\sqrt{2}-3\sqrt{3}+20\sqrt{2}-\left(3\sqrt{27}-6\sqrt{24}\right)$$

Combine 4\sqrt{2} and 20\sqrt{2} to get 24\sqrt{2}.

$$24\sqrt{2}-3\sqrt{3}-\left(3\sqrt{27}-6\sqrt{24}\right)$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$24\sqrt{2}-3\sqrt{3}-\left(3\times 3\sqrt{3}-6\sqrt{24}\right)$$

Multiply 3 and 3 to get 9.

$$24\sqrt{2}-3\sqrt{3}-\left(9\sqrt{3}-6\sqrt{24}\right)$$

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$24\sqrt{2}-3\sqrt{3}-\left(9\sqrt{3}-6\times 2\sqrt{6}\right)$$

Multiply -6 and 2 to get -12.

$$24\sqrt{2}-3\sqrt{3}-\left(9\sqrt{3}-12\sqrt{6}\right)$$

To find the opposite of 9\sqrt{3}-12\sqrt{6}, find the opposite of each term.

$$24\sqrt{2}-3\sqrt{3}-9\sqrt{3}-\left(-12\sqrt{6}\right)$$

Combine -3\sqrt{3} and -9\sqrt{3} to get -12\sqrt{3}.

$$24\sqrt{2}-12\sqrt{3}-\left(-12\sqrt{6}\right)$$

The opposite of -12\sqrt{6} is 12\sqrt{6}.

$$24\sqrt{2}-12\sqrt{3}+12\sqrt{6}$$