Solve (x+3)^2-(x-1)*(x+1)-3^2*(x+1)-1

$$( x + 3 ) ^ { 2 } - ( x - 1 ) \cdot ( x + 1 ) - 3 ^ { 2 } \cdot ( x + 1 ) - 1$$

Evaluate

$$-3x$$

Solution Steps

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

$$x^{2}+6x+9-\left(x-1\right)\left(x+1\right)-3^{2}\left(x+1\right)-1$$

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$x^{2}+6x+9-\left(x^{2}-1\right)-3^{2}\left(x+1\right)-1$$

To find the opposite of x^{2}-1, find the opposite of each term.

$$x^{2}+6x+9-x^{2}+1-3^{2}\left(x+1\right)-1$$

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

$$6x+9+1-3^{2}\left(x+1\right)-1$$

Add 9 and 1 to get 10.

$$6x+10-3^{2}\left(x+1\right)-1$$

Calculate 3 to the power of 2 and get 9.

$$6x+10-9\left(x+1\right)-1$$

Use the distributive property to multiply -9 by x+1.

$$6x+10-9x-9-1$$

Combine 6x and -9x to get -3x.

$$-3x+10-9-1$$

Subtract 9 from 10 to get 1.

$$-3x+1-1$$

Subtract 1 from 1 to get 0.

$$-3x$$

Expand

$$-3x$$

Solution Steps

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

$$x^{2}+6x+9-\left(x-1\right)\left(x+1\right)-3^{2}\left(x+1\right)-1$$

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$x^{2}+6x+9-\left(x^{2}-1\right)-3^{2}\left(x+1\right)-1$$

To find the opposite of x^{2}-1, find the opposite of each term.

$$x^{2}+6x+9-x^{2}+1-3^{2}\left(x+1\right)-1$$

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

$$6x+9+1-3^{2}\left(x+1\right)-1$$

Add 9 and 1 to get 10.

$$6x+10-3^{2}\left(x+1\right)-1$$

Calculate 3 to the power of 2 and get 9.

$$6x+10-9\left(x+1\right)-1$$

Use the distributive property to multiply -9 by x+1.

$$6x+10-9x-9-1$$

Combine 6x and -9x to get -3x.

$$-3x+10-9-1$$

Subtract 9 from 10 to get 1.

$$-3x+1-1$$

Subtract 1 from 1 to get 0.

$$-3x$$