Find x if (x + 1)^3 + (x - 2)^3 - 2x^2 (x - 1.5) = 3.

Step-by-step solution

1. 1

   Expand the two cubes with the binomial formula. Using $(x+b)^3=x^3+3bx^2+3b^2x+b^3$:

   $(x+1)^3=x^3+3x^2+3x+1$

   $(x-2)^3=x^3-6x^2+12x-8$

   The second one is where signs matter: $3(-2)^2=12$ is positive while $3(-2)=-6$ and $(-2)^3=-8$ are negative.

2. 2

   Add them.

   $(x+1)^3+(x-2)^3=2x^3-3x^2+15x-7$

3. 3

   Expand the subtracted product.

   $2x^2\left(x-1.5\right)=2x^3-3x^2$

   Note that $1.5=\tfrac32$, so $2x^2\cdot\tfrac32=3x^2$ — the decimal is there to disguise a clean fraction.

4. 4

   Subtract and watch the degree collapse.

   $\left(2x^3-3x^2+15x-7\right)-\left(2x^3-3x^2\right)=15x-7$

   Both the cubic and the quadratic terms cancel, which is the point of the problem: the coefficient $1.5$ was chosen precisely to kill the $x^2$ term.

5. 5

   Solve and verify.

   $15x-7=3\;\Longrightarrow\;15x=10\;\Longrightarrow\;x=\frac{2}{3}$

   Substituting back: $\left(\tfrac53\right)^3+\left(-\tfrac43\right)^3-2\cdot\tfrac49\left(\tfrac23-\tfrac32\right)=\tfrac{125}{27}-\tfrac{64}{27}+\tfrac{8}{9}\cdot\tfrac{5}{6}=\tfrac{61}{27}+\tfrac{20}{27}=\tfrac{81}{27}=3\;\checkmark$