Solve (1-x)^3+(1+x)^3/2=28

$$\frac { ( 1 - x ) ^ { 3 } + ( 1 + x ) ^ { 3 } } { 2 } = 28$$

Solve for x

$$x=3$$

$$x=-3$$

Steps Using Difference of Squares

Multiply both sides by 2.

$$\left(1-x\right)^{3}+\left(1+x\right)^{3}=28\times 2$$

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

$$1-3x+3x^{2}-x^{3}+\left(1+x\right)^{3}=28\times 2$$

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

$$1-3x+3x^{2}-x^{3}+1+3x+3x^{2}+x^{3}=28\times 2$$

Add 1 and 1 to get 2.

$$2-3x+3x^{2}-x^{3}+3x+3x^{2}+x^{3}=28\times 2$$

Combine -3x and 3x to get 0.

$$2+3x^{2}-x^{3}+3x^{2}+x^{3}=28\times 2$$

Combine 3x^{2} and 3x^{2} to get 6x^{2}.

$$2+6x^{2}-x^{3}+x^{3}=28\times 2$$

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

$$2+6x^{2}=28\times 2$$

Multiply 28 and 2 to get 56.

$$2+6x^{2}=56$$

Subtract 56 from both sides.

$$2+6x^{2}-56=0$$

Subtract 56 from 2 to get -54.

$$-54+6x^{2}=0$$

Divide both sides by 6.

$$-9+x^{2}=0$$

Consider -9+x^{2}. Rewrite -9+x^{2} as x^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-3\right)\left(x+3\right)=0$$

To find equation solutions, solve x-3=0 and x+3=0.

$$x=3 x=-3$$

Steps by Finding Square Root

Multiply both sides by 2.

$$\left(1-x\right)^{3}+\left(1+x\right)^{3}=28\times 2$$

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

$$1-3x+3x^{2}-x^{3}+\left(1+x\right)^{3}=28\times 2$$

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

$$1-3x+3x^{2}-x^{3}+1+3x+3x^{2}+x^{3}=28\times 2$$

Add 1 and 1 to get 2.

$$2-3x+3x^{2}-x^{3}+3x+3x^{2}+x^{3}=28\times 2$$

Combine -3x and 3x to get 0.

$$2+3x^{2}-x^{3}+3x^{2}+x^{3}=28\times 2$$

Combine 3x^{2} and 3x^{2} to get 6x^{2}.

$$2+6x^{2}-x^{3}+x^{3}=28\times 2$$

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

$$2+6x^{2}=28\times 2$$

Multiply 28 and 2 to get 56.

$$2+6x^{2}=56$$

Subtract 2 from both sides.

$$6x^{2}=56-2$$

Subtract 2 from 56 to get 54.

$$6x^{2}=54$$

Divide both sides by 6.

$$x^{2}=\frac{54}{6}$$

Divide 54 by 6 to get 9.

$$x^{2}=9$$

Take the square root of both sides of the equation.

$$x=3 x=-3$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$\left(1-x\right)^{3}+\left(1+x\right)^{3}=28\times 2$$

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

$$1-3x+3x^{2}-x^{3}+\left(1+x\right)^{3}=28\times 2$$

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

$$1-3x+3x^{2}-x^{3}+1+3x+3x^{2}+x^{3}=28\times 2$$

Add 1 and 1 to get 2.

$$2-3x+3x^{2}-x^{3}+3x+3x^{2}+x^{3}=28\times 2$$

Combine -3x and 3x to get 0.

$$2+3x^{2}-x^{3}+3x^{2}+x^{3}=28\times 2$$

Combine 3x^{2} and 3x^{2} to get 6x^{2}.

$$2+6x^{2}-x^{3}+x^{3}=28\times 2$$

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

$$2+6x^{2}=28\times 2$$

Multiply 28 and 2 to get 56.

$$2+6x^{2}=56$$

Subtract 56 from both sides.

$$2+6x^{2}-56=0$$

Subtract 56 from 2 to get -54.

$$-54+6x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$6x^{2}-54=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 0 for b, and -54 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 6\left(-54\right)}}{2\times 6}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 6\left(-54\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{0±\sqrt{-24\left(-54\right)}}{2\times 6}$$

Multiply -24 times -54.

$$x=\frac{0±\sqrt{1296}}{2\times 6}$$

Take the square root of 1296.

$$x=\frac{0±36}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{0±36}{12}$$

Now solve the equation x=\frac{0±36}{12} when ± is plus. Divide 36 by 12.

$$x=3$$

Now solve the equation x=\frac{0±36}{12} when ± is minus. Divide -36 by 12.

$$x=-3$$

The equation is now solved.

$$x=3 x=-3$$