Solve 1-a^6-3a^2+3a^4=

$$1 - a ^ { 6 } - 3 a ^ { 2 } + 3 a ^ { 4 } =$$

Factor

$$-\left(a-1\right)^{3}\left(a+1\right)^{3}$$

Solution Steps

To factor the expression, solve the equation where it equals to 0.

$$-a^{6}+3a^{4}-3a^{2}+1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient -1. List all candidates \frac{p}{q}.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$a=1$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide -a^{6}+3a^{4}-3a^{2}+1 by a-1 to get -a^{5}-a^{4}+2a^{3}+2a^{2}-a-1. To factor the result, solve the equation where it equals to 0.

$$-a^{5}-a^{4}+2a^{3}+2a^{2}-a-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient -1. List all candidates \frac{p}{q}.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$a=1$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide -a^{5}-a^{4}+2a^{3}+2a^{2}-a-1 by a-1 to get -a^{4}-2a^{3}+2a+1. To factor the result, solve the equation where it equals to 0.

$$-a^{4}-2a^{3}+2a+1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient -1. List all candidates \frac{p}{q}.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$a=-1$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide -a^{4}-2a^{3}+2a+1 by a+1 to get -a^{3}-a^{2}+a+1. To factor the result, solve the equation where it equals to 0.

$$-a^{3}-a^{2}+a+1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient -1. List all candidates \frac{p}{q}.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$a=-1$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide -a^{3}-a^{2}+a+1 by a+1 to get -a^{2}+1. To factor the result, solve the equation where it equals to 0.

$$-a^{2}+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute -1 for a, 0 for b, and 1 for c in the quadratic formula.

$$a=\frac{0±\sqrt{0^{2}-4\left(-1\right)\times 1}}{-2}$$

Do the calculations.

$$a=\frac{0±2}{-2}$$

Solve the equation -a^{2}+1=0 when ± is plus and when ± is minus.

$$a=1 a=-1$$

Rewrite the factored expression using the obtained roots.

$$\left(-a+1\right)\left(a-1\right)^{2}\left(a+1\right)^{3}$$

Evaluate

$$-\left(a^{2}-1\right)^{3}$$