Browse · MATH Print → jmc algebra intermediate Problem There exist constants c2, c1, and c0 such that x3+x2−5=(x−3)3+c2(x−3)2+c1(x−3)+c0.Find c22+c12+c02. Solution — click to reveal Let y=x−3. Then x=y+3, and x3+x2−5=(y+3)3+(y+3)2−5=y3+10y2+33y+31.Thus, c22+c12+c02=102+332+312=2150. Final answer 2150 ← Previous problem Next problem →