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Find (a³+b³+c³+d³)/(a²+b²+c²+d²) from Sum and Pair-Sum | IOQM

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Question
If a,b,c,da, b, c, d are real numbers such that a+b+c+d=20a + b + c + d = 20 and ab+ac+ad+bc+bd+cd=150ab + ac + ad + bc + bd + cd = 150, then find the value of
a3+b3+c3+d3a2+b2+c2+d2.\frac{a^3 + b^3 + c^3 + d^3}{a^2 + b^2 + c^2 + d^2}.
Solution
Answer: 5
Step 1: Find a2+b2+c2+d2a^2 + b^2 + c^2 + d^2. Using the identity
(a+b+c+d)2=a2+b2+c2+d2+2(ab+ac+ad+bc+bd+cd),(a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd),
 we obtain
a2+b2+c2+d2=(a+b+c+d)2−2(ab+ac+ad+bc+bd+cd)a^2 + b^2 + c^2 + d^2 = (a + b + c + d)^2 - 2(ab + ac + ad + bc + bd + cd)
202−2×150=400−300=100.20^2 - 2 \times 150 = 400 - 300 = 100.
 Step 2: Consider the sum of the squared pairwise differences. For four real numbers,
(a−b)2+(a−c)2+(a−d)2+(b−c)2+(b−d)2+(c−d)2(a - b)^2 + (a - c)^2 + (a - d)^2 + (b - c)^2 + (b - d)^2 + (c - d)^2
3(a2+b2+c2+d2)−2(ab+ac+ad+bc+bd+cd).3(a^2 + b^2 + c^2 + d^2) - 2(ab + ac + ad + bc + bd + cd).
 Substituting the known values,
=3×100−2×150=300−300=0.= 3 \times 100 - 2 \times 150 = 300 - 300 = 0.
 Step 3: A sum of squares of real numbers is zero only if every term is zero. Hence all pairwise differences vanish, so
a=b=c=d.a = b = c = d.
 Step 4: With a+b+c+d=20a + b + c + d = 20 and a=b=c=da = b = c = d, we get 4a=204a = 20, so a=b=c=d=5a = b = c = d = 5. Step 5: Evaluate the required expression.
a3+b3+c3+d3a2+b2+c2+d2=4×534×52=4×1254×25=500100=5.\frac{a^3 + b^3 + c^3 + d^3}{a^2 + b^2 + c^2 + d^2} = \frac{4 \times 5^3}{4 \times 5^2} = \frac{4 \times 125}{4 \times 25} = \frac{500}{100} = 5.
 Answer: 5
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