Question

Let a, b, c, d, e be five numbers satisfying the following conditions a + b + c + d + e=0; ABC+ABD+ABE+ACD+ACE+ ADE+BCD+BCE+BDE+CDE=33. Find the value of a^3 + b^3 + c^3 + d^3 + e^3. ​

Answer 1

Step-by-step explanation:

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Answer 2

Answer: 99

Step-by-step explanation:

Given: a, b, c, d, e be five numbers satisfying the following conditions a + b + c + d + e=0; ABC+ABD+ABE+ACD+ACE+ ADE+BCD+BCE+BDE+CDE=33.

To Find: The value of a^3 + b^3 + c^3 + d^3 + e^3.

Solution:

a + b + c + d + e=0   ....(i)

ABC+ABD+ABE+ACD+ACE+ ADE+BCD+BCE+BDE+CDE=33      .....(ii)

(a + b + c + d + e)³ = a³ + b³ + c³ + d³ + e³ + 3a²b + c + d + e + 3b²c + d + e + a + 3c²d + e + a + b + 3d²e + a + b + c + 3e²a + b + c + d + 6abc + abd + abc + ...

∵ a + b + c + d + e=0

b + c + d + e = -a

So, 3a²b + c + d + e = -3a³

Similarly, we get:

3b²c + a + d + e = -3b³

3c²d + a + b + e = -3c³

3d²e + a + b + c = -3d³

3e²a + b + c + d = -3e³

Now, from equation (i)

2a³ + b³ + c³  + d³ + e³ = 6abc + abd + abc + ...

⇒ a³ + b³ + c³  + d³ + e³ = 3abc + abd + abc + ...

⇒ a³ + b³ + c³  + d³ + e³ = 3 × 33 = 99

Hence, the required value of a³ + b³ + c³  + d³ + e³ is 99.

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