Solve (sqrt{10}+3)^2*(sqrt{10}-3)

$$( \sqrt { 10 } + 3 ) ^ { 2 } \times ( \sqrt { 10 } - 3 )$$

Evaluate

$$\sqrt{10}+3\approx 6.16227766$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{10}+3\right)^{2}.

$$\left(\left(\sqrt{10}\right)^{2}+6\sqrt{10}+9\right)\left(\sqrt{10}-3\right)$$

The square of \sqrt{10} is 10.

$$\left(10+6\sqrt{10}+9\right)\left(\sqrt{10}-3\right)$$

Add 10 and 9 to get 19.

$$\left(19+6\sqrt{10}\right)\left(\sqrt{10}-3\right)$$

Use the distributive property to multiply 19+6\sqrt{10} by \sqrt{10}-3 and combine like terms.

$$\sqrt{10}-57+6\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{10} is 10.

$$\sqrt{10}-57+6\times 10$$

Multiply 6 and 10 to get 60.

$$\sqrt{10}-57+60$$

Add -57 and 60 to get 3.

$$\sqrt{10}+3$$