Question

Factorize: $p^{3} (q-r)^{3} + q^{3} (r-p)^{3} + r^{3} (p-q)^{3}$

Answer 1

Answer:

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Answer 2

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✴ Required Answer:

✏ To factorise:

• p³(q-r)³ + q³(r-p)³ + r³(p-q)³

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✴ How to solve?

By using the corollary,

When a + b + c = 0

Then, a³ + b³ + c³=3abc

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✴ Solution:

We have p³(q-r)³ + q³(r-p)³ + r³(p-q)³ which can be written as, {p(q-r)}³ + {q(r-p)}³ + {r(p-q)}³

Taking this terms as a,b and c

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

This can be written as,

• a = pq-rp
• b = qr-pq
• c = rp-qr

Hence, here we can see that, pq-rp+qr-pq+rp-qr=0

• a + b + c = 0

By using corollary,

➝ a³+b³+c³ = 3abc

➝ {p(q-r)}³ + {q(r-p)}³ + {r(p-q)}³ = 3[p(q-r). q(r-p).r(p-q)]

➝ {p(q-r)}³ + {q(r-p)}³ + {r(p-q)}³ = 3pqr(q-r)(r-p)(p-q)

$\large{ \therefore{ \underline{ \underline{ \rm{ \pink{Hence \: factorised \: \dag}}}}}}$

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