Solve (x+1-b)^2+(x+3b)(x-3b)=2(x^2-4b^2)

$$( x + 1 - b ) ^ { 2 } + ( x + 3 b ) ( x - 3 b ) = 2 ( x ^ { 2 } - 4 b ^ { 2 } )$$

Solve for b

$$b=\frac{2x+1}{2\left(x+1\right)}$$

$$x\neq -1$$

Steps for Solving Linear Equation

Square x+1-b.

$$x^{2}-2bx+2x+b^{2}-2b+1+\left(x+3b\right)\left(x-3b\right)=2\left(x^{2}-4b^{2}\right)$$

Consider \left(x+3b\right)\left(x-3b\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x^{2}-2bx+2x+b^{2}-2b+1+x^{2}-\left(3b\right)^{2}=2\left(x^{2}-4b^{2}\right)$$

Expand \left(3b\right)^{2}.

$$x^{2}-2bx+2x+b^{2}-2b+1+x^{2}-3^{2}b^{2}=2\left(x^{2}-4b^{2}\right)$$

Calculate 3 to the power of 2 and get 9.

$$x^{2}-2bx+2x+b^{2}-2b+1+x^{2}-9b^{2}=2\left(x^{2}-4b^{2}\right)$$

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

$$2x^{2}-2bx+2x+b^{2}-2b+1-9b^{2}=2\left(x^{2}-4b^{2}\right)$$

Combine b^{2} and -9b^{2} to get -8b^{2}.

$$2x^{2}-2bx+2x-8b^{2}-2b+1=2\left(x^{2}-4b^{2}\right)$$

Use the distributive property to multiply 2 by x^{2}-4b^{2}.

$$2x^{2}-2bx+2x-8b^{2}-2b+1=2x^{2}-8b^{2}$$

Add 8b^{2} to both sides.

$$2x^{2}-2bx+2x-8b^{2}-2b+1+8b^{2}=2x^{2}$$

Combine -8b^{2} and 8b^{2} to get 0.

$$2x^{2}-2bx+2x-2b+1=2x^{2}$$

Subtract 2x^{2} from both sides.

$$-2bx+2x-2b+1=2x^{2}-2x^{2}$$

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

$$-2bx+2x-2b+1=0$$

Subtract 2x from both sides. Anything subtracted from zero gives its negation.

$$-2bx-2b+1=-2x$$

Subtract 1 from both sides.

$$-2bx-2b=-2x-1$$

Combine all terms containing b.

$$\left(-2x-2\right)b=-2x-1$$

Divide both sides by -2x-2.

$$\frac{\left(-2x-2\right)b}{-2x-2}=\frac{-2x-1}{-2x-2}$$

Dividing by -2x-2 undoes the multiplication by -2x-2.

$$b=\frac{-2x-1}{-2x-2}$$

Divide -2x-1 by -2x-2.

$$b=\frac{2x+1}{2\left(x+1\right)}$$

Solve for x

$$x=-\frac{1-2b}{2\left(1-b\right)}$$

$$b\neq 1$$

Steps for Solving Linear Equation

Square x+1-b.

$$x^{2}-2bx+2x+b^{2}-2b+1+\left(x+3b\right)\left(x-3b\right)=2\left(x^{2}-4b^{2}\right)$$

Consider \left(x+3b\right)\left(x-3b\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x^{2}-2bx+2x+b^{2}-2b+1+x^{2}-\left(3b\right)^{2}=2\left(x^{2}-4b^{2}\right)$$

Expand \left(3b\right)^{2}.

$$x^{2}-2bx+2x+b^{2}-2b+1+x^{2}-3^{2}b^{2}=2\left(x^{2}-4b^{2}\right)$$

Calculate 3 to the power of 2 and get 9.

$$x^{2}-2bx+2x+b^{2}-2b+1+x^{2}-9b^{2}=2\left(x^{2}-4b^{2}\right)$$

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

$$2x^{2}-2bx+2x+b^{2}-2b+1-9b^{2}=2\left(x^{2}-4b^{2}\right)$$

Combine b^{2} and -9b^{2} to get -8b^{2}.

$$2x^{2}-2bx+2x-8b^{2}-2b+1=2\left(x^{2}-4b^{2}\right)$$

Use the distributive property to multiply 2 by x^{2}-4b^{2}.

$$2x^{2}-2bx+2x-8b^{2}-2b+1=2x^{2}-8b^{2}$$

Subtract 2x^{2} from both sides.

$$2x^{2}-2bx+2x-8b^{2}-2b+1-2x^{2}=-8b^{2}$$

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

$$-2bx+2x-8b^{2}-2b+1=-8b^{2}$$

Add 8b^{2} to both sides.

$$-2bx+2x-2b+1=-8b^{2}+8b^{2}$$

Combine -8b^{2} and 8b^{2} to get 0.

$$-2bx+2x-2b+1=0$$

Add 2b to both sides. Anything plus zero gives itself.

$$-2bx+2x+1=2b$$

Subtract 1 from both sides.

$$-2bx+2x=2b-1$$

Combine all terms containing x.

$$\left(-2b+2\right)x=2b-1$$

The equation is in standard form.

$$\left(2-2b\right)x=2b-1$$

Divide both sides by -2b+2.

$$\frac{\left(2-2b\right)x}{2-2b}=\frac{2b-1}{2-2b}$$

Dividing by -2b+2 undoes the multiplication by -2b+2.

$$x=\frac{2b-1}{2-2b}$$

Divide 2b-1 by -2b+2.

$$x=\frac{2b-1}{2\left(1-b\right)}$$