Solve (9-x)(9+x)+(x+3)(x+3)

$$( 9 - x ) ( 9 + x ) + ( x + 3 ) ( x + 3 )$$

Evaluate

$$6\left(x+15\right)$$

Solution Steps

Multiply x+3 and x+3 to get \left(x+3\right)^{2}.

$$\left(9-x\right)\left(9+x\right)+\left(x+3\right)^{2}$$

Consider \left(9-x\right)\left(9+x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$9^{2}-x^{2}+\left(x+3\right)^{2}$$

Calculate 9 to the power of 2 and get 81.

$$81-x^{2}+\left(x+3\right)^{2}$$

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

$$81-x^{2}+x^{2}+6x+9$$

Combine -x^{2} and x^{2} to get 0.

$$81+6x+9$$

Add 81 and 9 to get 90.

$$90+6x$$

Expand

$$6x+90$$

Solution Steps

Multiply x+3 and x+3 to get \left(x+3\right)^{2}.

$$\left(9-x\right)\left(9+x\right)+\left(x+3\right)^{2}$$

Consider \left(9-x\right)\left(9+x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$9^{2}-x^{2}+\left(x+3\right)^{2}$$

Calculate 9 to the power of 2 and get 81.

$$81-x^{2}+\left(x+3\right)^{2}$$

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

$$81-x^{2}+x^{2}+6x+9$$

Combine -x^{2} and x^{2} to get 0.

$$81+6x+9$$

Add 81 and 9 to get 90.

$$90+6x$$