Solve 27r^12s^6+r^3=

$$27 r ^ { 12 } s ^ { 6 } + r ^ { 3 } =$$

Factor

$$r^{3}\left(3s^{2}r^{3}+1\right)\left(9s^{4}r^{6}-3s^{2}r^{3}+1\right)$$

Solution Steps

Factor out r^{3}.

$$r^{3}\left(27r^{9}s^{6}+1\right)$$

Consider 27r^{9}s^{6}+1. Consider 27r^{9}s^{6}+1 as a polynomial over variable r.

$$27s^{6}r^{9}+1$$

Find one factor of the form ks^{m}r^{n}+p, where ks^{m}r^{n} divides the monomial with the highest power 27s^{6}r^{9} and p divides the constant factor 1. One such factor is 3s^{2}r^{3}+1. Factor the polynomial by dividing it by this factor.

$$\left(3s^{2}r^{3}+1\right)\left(9s^{4}r^{6}-3s^{2}r^{3}+1\right)$$

Rewrite the complete factored expression.

$$r^{3}\left(3s^{2}r^{3}+1\right)\left(9s^{4}r^{6}-3s^{2}r^{3}+1\right)$$

Evaluate

$$r^{3}+27s^{6}r^{12}$$