Solve (sqrt{3}+27)(sqrt{3}+sqrt{7})^2

$$( \sqrt { 3 } + 27 ) ( \sqrt { 3 } + \sqrt { 7 } ) ^ { 2 }$$

Evaluate

$$6\sqrt{7}+10\sqrt{3}+54\sqrt{21}+270\approx 550.65410347$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+\sqrt{7}\right)^{2}.

$$\left(\sqrt{3}+27\right)\left(\left(\sqrt{3}\right)^{2}+2\sqrt{3}\sqrt{7}+\left(\sqrt{7}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$\left(\sqrt{3}+27\right)\left(3+2\sqrt{3}\sqrt{7}+\left(\sqrt{7}\right)^{2}\right)$$

To multiply \sqrt{3} and \sqrt{7}, multiply the numbers under the square root.

$$\left(\sqrt{3}+27\right)\left(3+2\sqrt{21}+\left(\sqrt{7}\right)^{2}\right)$$

The square of \sqrt{7} is 7.

$$\left(\sqrt{3}+27\right)\left(3+2\sqrt{21}+7\right)$$

Add 3 and 7 to get 10.

$$\left(\sqrt{3}+27\right)\left(10+2\sqrt{21}\right)$$

Use the distributive property to multiply \sqrt{3}+27 by 10+2\sqrt{21}.

$$10\sqrt{3}+2\sqrt{3}\sqrt{21}+270+54\sqrt{21}$$

Factor 21=3\times 7. Rewrite the square root of the product \sqrt{3\times 7} as the product of square roots \sqrt{3}\sqrt{7}.

$$10\sqrt{3}+2\sqrt{3}\sqrt{3}\sqrt{7}+270+54\sqrt{21}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$10\sqrt{3}+2\times 3\sqrt{7}+270+54\sqrt{21}$$

Multiply 2 and 3 to get 6.

$$10\sqrt{3}+6\sqrt{7}+270+54\sqrt{21}$$