Question

factorise p^3(q - r)^3 + q^3(r - p)^3 + r^3(p - q)^3
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Answer 1

Answer:

Hii

Step-by-step explanation:

3pqr(q - r)(r - p)(p - q)

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Given :

To factorize :

p³ (q - r)³ + q³ (r - p)³+r³(p - q)³

Solution :

p³ (q - r)³ + q³ (r - p)³+r³(p - q)³

⇒ [p (q - r)]³+[q (r - p)]³+ [r (p - q)]³

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if,

a = p (q - r) , b = q (r - p) , c = r (p - q)

By adding a , b & c,

We get,

⇒ a + b + c = p (q - r) + q (r - p) + (p - q) = (pq - pr) + (qr - pq) + (pr - qr) = pq - pq + qr - qr + pr - pr = 0

We know that,

If a + b + c = 0,

then,

a³ + b³ + c³ = 3abc (Identity)

So,

By substituting,

a = p (q - r) , b = q (r - p) , c = r (p - q)

We get,

⇒ [p(q - r)]³ + [q(r - q)]³ + [r(p - q)]³ = 3pqr (q - r)(r - p)(p - q)

⇒p³ (q - r)³ + q³ (r - p)³ + r³ (p - q)³ = 3pqr (q - r)(r - p)(p - q)

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The factorized form of p³ (q - r)³ + q³ (r - p)³ + r³ (p - q)³ is 3pqr (q - r)(r - p)(p - q)

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Answer 2

Answer:

3pqr(p - q)(q - r)(r - p)

Step-by-step explanation:

Given Equation is p³(q - r)³ + q³(r - p)³ + r³(p - q)³

⇒ [p(q - r)]³ + [q(r - p)]³ + [r(p - q)]³

Let a = p(q - r), b = q(r - p), c = r(p - q)

∴ a + b + c = p(q - r) + q(r - p) + r(p - q)

= pq - pr + qr - qp + rp - rq

= 0

We know that when a + b + c = 0, then a³ + b³ + c³ = 3abc

⇒ 3(pq - qr)(qr - qp)(rp - rq)

⇒ 3pqr(p - q)(q - r)(r - p)

Hope it helps!