If a + b + c = 6, a2 + b2 + c2 = 32, and a3 + b3 + c3 = 189, then the value of 4abc is:
Answer & explanation
Correct answer: option 1
a + b + c = 6
a2 + b2 + c2 = 32
a3 + b3 + c3 = 189
= (a + b + c)2 - 2(ab + bc + ca) = 32
= 62 - 2(ab + bc + ca) = 32
= -2(ab + bc + ca) = 32 - 36
= -2(ab + bc + ca) = -4
= (ab + bc + ca) = 2 ....(a)
a3 + b3 + c3 = 189
= (a + b + c) (a2 + b2 + c2 - (ab + bc + ca)) + 3abc = 189
= 6 × (32 - 2) + 3abc = 189 (From a)
= 6 × 30 + 3abc = 189
= 3abc = 189 - 180
= 4abc = \(\frac{9 × 4}{ 3}\)
= 4abc = 12