a=2x-1,b=2x-2 c=3-4x , a^3+b^3+c^3=?
asked by guest on Mar 29, 2025 at 5:04 pm
MathBot Answer:
a3+b3+c3=(3−4x)3+(2x−2)3+(2x−1)3{a}^{3} + {b}^{3} + {c}^{3} = \left(3 - 4 x\right)^{3} + \left(2 x - 2\right)^{3} + \left(2 x - 1\right)^{3}a3+b3+c3=(3−4x)3+(2x−2)3+(2x−1)3
a3+b3+c3=b3+c3+(2x−1)3=c3+(2x−2)3+(2x−1)3=(3−4x)3+(2x−2)3+(2x−1)3=(3−4x)3+(2x−1)3+(2x−2)3=(−4x+3)3+(2x−1)3+(2x−2)3=(−4x+3)3+(2x−1)3+8x3−24x2+24x−8=(−4x+3)3+8x3−12x2+6x−1+8x3−24x2+24x−8=8x3−12x2+6x−1+8x3−24x2+24x−8−64x3+144x2−108x+27=(3−4x)3+(2x−2)3+(2x−1)3\begin{aligned}a^{3} + b^{3} + c^{3}&=b^{3} + c^{3} + \left(2 x - 1\right)^{3}\\&=c^{3} + \left(2 x - 2\right)^{3} + \left(2 x - 1\right)^{3}\\&=\left(3 - 4 x\right)^{3} + \left(2 x - 2\right)^{3} + \left(2 x - 1\right)^{3}\\&=\left(3 - 4 x\right)^{3} + \left(2 x - 1\right)^{3} + \left(2 x - 2\right)^{3}\\&=\left(- 4 x + 3\right)^{3} + \left(2 x - 1\right)^{3} + \left(2 x - 2\right)^{3}\\&=\left(- 4 x + 3\right)^{3} + \left(2 x - 1\right)^{3} + 8 x^{3} - 24 x^{2} + 24 x - 8\\&=\left(- 4 x + 3\right)^{3} + 8 x^{3} - 12 x^{2} + 6 x - 1 + 8 x^{3} - 24 x^{2} + 24 x - 8\\&=8 x^{3} - 12 x^{2} + 6 x - 1 + 8 x^{3} - 24 x^{2} + 24 x - 8 - 64 x^{3} + 144 x^{2} - 108 x + 27\\&=\left(3 - 4 x\right)^{3} + \left(2 x - 2\right)^{3} + \left(2 x - 1\right)^{3}\end{aligned}a3+b3+c3=b3+c3+(2x−1)3=c3+(2x−2)3+(2x−1)3=(3−4x)3+(2x−2)3+(2x−1)3=(3−4x)3+(2x−1)3+(2x−2)3=(−4x+3)3+(2x−1)3+(2x−2)3=(−4x+3)3+(2x−1)3+8x3−24x2+24x−8=(−4x+3)3+8x3−12x2+6x−1+8x3−24x2+24x−8=8x3−12x2+6x−1+8x3−24x2+24x−8−64x3+144x2−108x+27=(3−4x)3+(2x−2)3+(2x−1)3
