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

JEE Maths question with a full step-by-step solution.

Question

a, b, c, d

are real numbers such that

a + b + c + d = 20

and

ab + ac + ad + bc + bd + cd = 150

, then find the value of

\frac{a^3 + b^3 + c^3 + d^3}{a^2 + b^2 + c^2 + d^2}.

Solution

Answer: 5

Step 1: Find

a^2 + b^2 + c^2 + d^2

. Using the identity

(a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd),

we obtain

a^2 + b^2 + c^2 + d^2 = (a + b + c + d)^2 - 2(ab + ac + ad + bc + bd + cd)

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

3(a^2 + b^2 + c^2 + d^2) - 2(ab + ac + ad + bc + bd + cd).

Substituting the known values,

= 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.

Step 4: With

a + b + c + d = 20

and

a = b = c = d

, we get

4a = 20

, so

a = b = c = d = 5

. Step 5: Evaluate the required expression.

\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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