Solve (x-7)^3=(9-x)^2

$$( x - 7 ) ^ { 3 } = ( 9 - x ) ^ { 2 }$$

Solve for x (complex solution)

$$x=7-2i$$

$$x=8$$

$$x=7+2i$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-7\right)^{3}.

$$x^{3}-21x^{2}+147x-343=\left(9-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-x\right)^{2}.

$$x^{3}-21x^{2}+147x-343=81-18x+x^{2}$$

Subtract 81 from both sides.

$$x^{3}-21x^{2}+147x-343-81=-18x+x^{2}$$

Subtract 81 from -343 to get -424.

$$x^{3}-21x^{2}+147x-424=-18x+x^{2}$$

Add 18x to both sides.

$$x^{3}-21x^{2}+147x-424+18x=x^{2}$$

Combine 147x and 18x to get 165x.

$$x^{3}-21x^{2}+165x-424=x^{2}$$

Subtract x^{2} from both sides.

$$x^{3}-21x^{2}+165x-424-x^{2}=0$$

Combine -21x^{2} and -x^{2} to get -22x^{2}.

$$x^{3}-22x^{2}+165x-424=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -424 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±424,±212,±106,±53,±8,±4,±2,±1$$

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.

$$x=8$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-22x^{2}+165x-424 by x-8 to get x^{2}-14x+53. Solve the equation where the result equals to 0.

$$x^{2}-14x+53=0$$

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, -14 for b, and 53 for c in the quadratic formula.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 1\times 53}}{2}$$

Do the calculations.

$$x=\frac{14±\sqrt{-16}}{2}$$

Solve the equation x^{2}-14x+53=0 when ± is plus and when ± is minus.

$$x=7-2i x=7+2i$$

List all found solutions.

$$x=8 x=7-2i x=7+2i$$

Solve for x

$$x=8$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-7\right)^{3}.

$$x^{3}-21x^{2}+147x-343=\left(9-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-x\right)^{2}.

$$x^{3}-21x^{2}+147x-343=81-18x+x^{2}$$

Subtract 81 from both sides.

$$x^{3}-21x^{2}+147x-343-81=-18x+x^{2}$$

Subtract 81 from -343 to get -424.

$$x^{3}-21x^{2}+147x-424=-18x+x^{2}$$

Add 18x to both sides.

$$x^{3}-21x^{2}+147x-424+18x=x^{2}$$

Combine 147x and 18x to get 165x.

$$x^{3}-21x^{2}+165x-424=x^{2}$$

Subtract x^{2} from both sides.

$$x^{3}-21x^{2}+165x-424-x^{2}=0$$

Combine -21x^{2} and -x^{2} to get -22x^{2}.

$$x^{3}-22x^{2}+165x-424=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -424 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±424,±212,±106,±53,±8,±4,±2,±1$$

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.

$$x=8$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-22x^{2}+165x-424 by x-8 to get x^{2}-14x+53. Solve the equation where the result equals to 0.

$$x^{2}-14x+53=0$$

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, -14 for b, and 53 for c in the quadratic formula.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 1\times 53}}{2}$$

Do the calculations.

$$x=\frac{14±\sqrt{-16}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=8$$