If a + b + c = 5, a2 + b2 + c2 = 27 and a3 + b3 + c3 = 125, then the value of 4abc is:
Show answer & explanation
Correct answer: C
Explanation
Correct answer is option 3 i.e. -20
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Concept |
Understanding |
Application |
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Formula: x3+y3+z3-3xyz = (x+y+z) (x2+y2+z2-xy-yz-zx) and (x+y+z)2 = (x2+y2+z2+2xy+2yz+2xz) |
Given: a + b + c = 5, a2 + b2 + c2 = 27, and a3 + b3 + c3 = 125 4abc = ? First we need to find the value of (ab+bc+ca) |
(a+b+c)2 = (a2+b2+c2+2ab+2bc+2ca) 25 = (27 + 2(ab+bc+ca) ab+bc+ca = (25 – 27)/2 = -1 Now a3+b3+c3-3abc = (a+b+c) (a2+b2+c2-ab-bc-ca) a3+b3+c3-3abc = (a+b+c) ((a+b+c)2-3(ab+bc+ca)) 125 – 3abc = 5 × (25 – 3 × -1) 3abc = 125 - 140 3abc = -15 abc = -5 Hence, 4abc = -20 |