Solve (x-3)^2-25x^2+6x=(2x-1)^3-(1+2x)^3+11

$$( x - 3 ) ^ { 2 } - 25 x ^ { 2 } + 6 x = ( 2 x - 1 ) ^ { 3 } - ( 1 + 2 x ) ^ { 3 } + 11$$

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Solution Steps

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

$$x^{2}-6x+9-25x^{2}+6x=\left(2x-1\right)^{3}-\left(1+2x\right)^{3}+11$$

Combine x^{2} and -25x^{2} to get -24x^{2}.

$$-24x^{2}-6x+9+6x=\left(2x-1\right)^{3}-\left(1+2x\right)^{3}+11$$

Combine -6x and 6x to get 0.

$$-24x^{2}+9=\left(2x-1\right)^{3}-\left(1+2x\right)^{3}+11$$

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

$$-24x^{2}+9=8x^{3}-12x^{2}+6x-1-\left(1+2x\right)^{3}+11$$

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

$$-24x^{2}+9=8x^{3}-12x^{2}+6x-1-\left(1+6x+12x^{2}+8x^{3}\right)+11$$

To find the opposite of 1+6x+12x^{2}+8x^{3}, find the opposite of each term.

$$-24x^{2}+9=8x^{3}-12x^{2}+6x-1-1-6x-12x^{2}-8x^{3}+11$$

Subtract 1 from -1 to get -2.

$$-24x^{2}+9=8x^{3}-12x^{2}+6x-2-6x-12x^{2}-8x^{3}+11$$

Combine 6x and -6x to get 0.

$$-24x^{2}+9=8x^{3}-12x^{2}-2-12x^{2}-8x^{3}+11$$

Combine -12x^{2} and -12x^{2} to get -24x^{2}.

$$-24x^{2}+9=8x^{3}-24x^{2}-2-8x^{3}+11$$

Combine 8x^{3} and -8x^{3} to get 0.

$$-24x^{2}+9=-24x^{2}-2+11$$

Add -2 and 11 to get 9.

$$-24x^{2}+9=-24x^{2}+9$$

Add 24x^{2} to both sides.

$$-24x^{2}+9+24x^{2}=9$$

Combine -24x^{2} and 24x^{2} to get 0.

$$9=9$$

Compare 9 and 9.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

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

$$x^{2}-6x+9-25x^{2}+6x=\left(2x-1\right)^{3}-\left(1+2x\right)^{3}+11$$

Combine x^{2} and -25x^{2} to get -24x^{2}.

$$-24x^{2}-6x+9+6x=\left(2x-1\right)^{3}-\left(1+2x\right)^{3}+11$$

Combine -6x and 6x to get 0.

$$-24x^{2}+9=\left(2x-1\right)^{3}-\left(1+2x\right)^{3}+11$$

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

$$-24x^{2}+9=8x^{3}-12x^{2}+6x-1-\left(1+2x\right)^{3}+11$$

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

$$-24x^{2}+9=8x^{3}-12x^{2}+6x-1-\left(1+6x+12x^{2}+8x^{3}\right)+11$$

To find the opposite of 1+6x+12x^{2}+8x^{3}, find the opposite of each term.

$$-24x^{2}+9=8x^{3}-12x^{2}+6x-1-1-6x-12x^{2}-8x^{3}+11$$

Subtract 1 from -1 to get -2.

$$-24x^{2}+9=8x^{3}-12x^{2}+6x-2-6x-12x^{2}-8x^{3}+11$$

Combine 6x and -6x to get 0.

$$-24x^{2}+9=8x^{3}-12x^{2}-2-12x^{2}-8x^{3}+11$$

Combine -12x^{2} and -12x^{2} to get -24x^{2}.

$$-24x^{2}+9=8x^{3}-24x^{2}-2-8x^{3}+11$$

Combine 8x^{3} and -8x^{3} to get 0.

$$-24x^{2}+9=-24x^{2}-2+11$$

Add -2 and 11 to get 9.

$$-24x^{2}+9=-24x^{2}+9$$

Add 24x^{2} to both sides.

$$-24x^{2}+9+24x^{2}=9$$

Combine -24x^{2} and 24x^{2} to get 0.

$$9=9$$

Compare 9 and 9.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$