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If p, q, r are all real numbers, then (p – q)3 + (q – r)3 + (r – p)3 is equal to       (SSC Sub. Inspector 2016)
  • a)
    0
  • b)
    3 (p – q) (q – r) (r – p)
  • c)
    (p – q) (q – r) (r – p)
  • d)
    1
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
If p, q, r are all real numbers, then (p – q)3 + (q – r)3 ...
If a + b + c = 0 then a3 + b3 + c3 = 3abc
let, a = (p – q), b = (q – r), c - (r – p)
than, a + b + c = p – q + q – r + r – p = 0
∴ (p – q)3 + (q – r)2 + (r – p)3
= 3(p – q) (q – r) (r – p) = 0
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Community Answer
If p, q, r are all real numbers, then (p – q)3 + (q – r)3 ...
Explanation:
The given expression can be simplified by expanding the cube of binomials and then combining like terms.
1. Expand the cube terms:
(p - q)^3 = p^3 - 3p^2q + 3pq^2 - q^3
(q - r)^3 = q^3 - 3q^2r + 3qr^2 - r^3
(r - p)^3 = r^3 - 3r^2p + 3rp^2 - p^3
2. Add the expanded terms:
(p - q)^3 + (q - r)^3 + (r - p)^3
= p^3 - 3p^2q + 3pq^2 - q^3 + q^3 - 3q^2r + 3qr^2 - r^3 + r^3 - 3r^2p + 3rp^2 - p^3
= p^3 - q^3 - r^3 - 3p^2q + 3pq^2 - 3q^2r + 3qr^2 - 3r^2p + 3rp^2
3. Factor out (p - q)(q - r)(r - p):
= (p - q)(-p^2 + q^2 - r^2 + pr - qr + pq)
= (p - q)(q - r)(r - p)
Therefore, the expression (p - q)^3 + (q - r)^3 + (r - p)^3 simplifies to (p - q)(q - r)(r - p), which is equal to option 'c'.
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If p, q, r are all real numbers, then (p – q)3 + (q – r)3 + (r – p)3 is equal to (SSC Sub. Inspector 2016)a)0b)3 (p – q) (q – r) (r – p)c)(p – q) (q – r) (r – p)d)1Correct answer is option 'C'. Can you explain this answer?
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