Solve y=(x^{2}+1)^3*(x^2-x^3)^2

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

Solve for y

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

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x^{2}+1\right)^{3}.

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

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

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

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

$$y=\left(x^{6}+3x^{4}+3x^{2}+1\right)\left(x^{2}-x^{3}\right)^{2}$$

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

$$y=\left(x^{6}+3x^{4}+3x^{2}+1\right)\left(\left(x^{2}\right)^{2}-2x^{2}x^{3}+\left(x^{3}\right)^{2}\right)$$

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

$$y=\left(x^{6}+3x^{4}+3x^{2}+1\right)\left(x^{4}-2x^{2}x^{3}+\left(x^{3}\right)^{2}\right)$$

To multiply powers of the same base, add their exponents. Add 2 and 3 to get 5.

$$y=\left(x^{6}+3x^{4}+3x^{2}+1\right)\left(x^{4}-2x^{5}+\left(x^{3}\right)^{2}\right)$$

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

$$y=\left(x^{6}+3x^{4}+3x^{2}+1\right)\left(x^{4}-2x^{5}+x^{6}\right)$$

Use the distributive property to multiply x^{6}+3x^{4}+3x^{2}+1 by x^{4}-2x^{5}+x^{6} and combine like terms.

$$y=4x^{10}-2x^{11}+x^{12}+6x^{8}-6x^{9}+4x^{6}-6x^{7}+x^{4}-2x^{5}$$