Solve -f(x)=(x^2-3)^3(x+1)^2

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

Solve for f

$$f=-\frac{\left(x+1\right)^{2}\left(x^{2}-3\right)^{3}}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$\left(-f\right)x=\left(\left(x^{2}\right)^{3}-9\left(x^{2}\right)^{2}+27x^{2}-27\right)\left(x+1\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.

$$\left(-f\right)x=\left(x^{6}-9\left(x^{2}\right)^{2}+27x^{2}-27\right)\left(x+1\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\left(-f\right)x=\left(x^{6}-9x^{4}+27x^{2}-27\right)\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$\left(-f\right)x=\left(x^{6}-9x^{4}+27x^{2}-27\right)\left(x^{2}+2x+1\right)$$

Use the distributive property to multiply x^{6}-9x^{4}+27x^{2}-27 by x^{2}+2x+1 and combine like terms.

$$\left(-f\right)x=x^{8}+2x^{7}-8x^{6}-18x^{5}+18x^{4}+54x^{3}-54x-27$$

Reorder the terms.

$$-fx=x^{8}+2x^{7}-8x^{6}-18x^{5}+18x^{4}+54x^{3}-54x-27$$

The equation is in standard form.

$$\left(-x\right)f=x^{8}+2x^{7}-8x^{6}-18x^{5}+18x^{4}+54x^{3}-54x-27$$

Divide both sides by -x.

$$\frac{\left(-x\right)f}{-x}=\frac{\left(x+1\right)^{2}\left(x^{2}-3\right)^{3}}{-x}$$

Dividing by -x undoes the multiplication by -x.

$$f=\frac{\left(x+1\right)^{2}\left(x^{2}-3\right)^{3}}{-x}$$

Divide \left(-3+x^{2}\right)^{3}\left(1+x\right)^{2} by -x.

$$f=-\frac{\left(x+1\right)^{2}\left(x^{2}-3\right)^{3}}{x}$$