Solve {l}{(4-3z)^{3}+z(14+27z^{2})=108z^2+77}{(5x+2)^3-25x(5x^2-4)=150x^2+21}

$$\left. \begin{array} { l } { ( 4 - 3 z ) ^ { 3 } + z ( 14 + 27 z ^ { 2 } ) = 108 z ^ { 2 } + 77 } \\ { ( 5 x + 2 ) ^ { 3 } - 25 x ( 5 x ^ { 2 } - 4 ) = 150 x ^ { 2 } + 21 } \end{array} \right.$$

Solve for z, x

$$x=\frac{13}{160}=0.08125$$

$$z=-\frac{1}{10}=-0.1$$

Solution Steps

Consider the first equation. Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(4-3z\right)^{3}.

$$64-144z+108z^{2}-27z^{3}+z\left(14+27z^{2}\right)=108z^{2}+77$$

Use the distributive property to multiply z by 14+27z^{2}.

$$64-144z+108z^{2}-27z^{3}+14z+27z^{3}=108z^{2}+77$$

Combine -144z and 14z to get -130z.

$$64-130z+108z^{2}-27z^{3}+27z^{3}=108z^{2}+77$$

Combine -27z^{3} and 27z^{3} to get 0.

$$64-130z+108z^{2}=108z^{2}+77$$

Subtract 108z^{2} from both sides.

$$64-130z+108z^{2}-108z^{2}=77$$

Combine 108z^{2} and -108z^{2} to get 0.

$$64-130z=77$$

Subtract 64 from both sides.

$$-130z=77-64$$

Subtract 64 from 77 to get 13.

$$-130z=13$$

Divide both sides by -130.

$$z=\frac{13}{-130}$$

Reduce the fraction \frac{13}{-130} to lowest terms by extracting and canceling out 13.

$$z=-\frac{1}{10}$$

Consider the second equation. Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(5x+2\right)^{3}.

$$125x^{3}+150x^{2}+60x+8-25x\left(5x^{2}-4\right)=150x^{2}+21$$

Subtract 150x^{2} from both sides.

$$125x^{3}+150x^{2}+60x+8-25x\left(5x^{2}-4\right)-150x^{2}=21$$

Use the distributive property to multiply -25x by 5x^{2}-4.

$$125x^{3}+150x^{2}+60x+8-125x^{3}+100x-150x^{2}=21$$

Combine 125x^{3} and -125x^{3} to get 0.

$$150x^{2}+60x+8+100x-150x^{2}=21$$

Combine 60x and 100x to get 160x.

$$150x^{2}+160x+8-150x^{2}=21$$

Combine 150x^{2} and -150x^{2} to get 0.

$$160x+8=21$$

Subtract 8 from both sides.

$$160x=21-8$$

Subtract 8 from 21 to get 13.

$$160x=13$$

Divide both sides by 160.

$$x=\frac{13}{160}$$

The system is now solved.

$$z=-\frac{1}{10} x=\frac{13}{160}$$