Question

For question refer to attachment​

Answer 1

Proper question:-

$\dfrac{p^{3}(q-r)^{3}-q^{3}(p-r)^{3}-r^{3}(q-p)^{3}}{(p-q)(r-q)(p-r)}$ is equal to (where $p,q,r\neq 0$ and $p\neq q$, $q\neq r$, $r\neq p$)

• $pqr$
• $3pqr$
• $3(p+q+r)$
• $3$

Solution:-

The numerator is given as a sum of cubes. To simplify this fraction we choose factorization method.

We see that the sum of each term without exponents is zero. This means we can use the following identity.

$\boxed{\bold{a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)}}$

Since $a+b+c=0$, this leads to $a^{3}+b^{3}+c^{3}-3abc=0$, which then leads to $a^{3}+b^{3}+c^{3}=3abc$.

Now the sum of cubes is equal to three times the product of three numbers! Using this method, we can simplify the given fraction.

Given:-

$\dfrac{p^{3}(q-r)^{3}-q^{3}(p-r)^{3}-r^{3}(q-p)^{3}}{(p-q)(r-q)(p-r)}$

$=\dfrac{(pq-rp)^{3}-(pq-qr)^{3}-(qr-rp)^{3}}{(p-q)(r-q)(p-r)}$

$=\dfrac{(pq-rp)^{3}+(qr-pq)^{3}+(rp-qr)^{3}}{(p-q)(q-r)(r-p)}$

We know that $a^{3}+b^{3}+c^{3}=3abc$ if $a+b+c=0$.

Here, $(pq-rp)+(qr-pq)+(rp-qr)=0$, so we can simplify this as three times the product of three numbers!

Given:-

$\dfrac{3(pq-rp)(qr-pq)(rp-qr)}{(p-q)(q-r)(r-p)}$

$=\dfrac{3pqr(q-r)(r-p)(p-q)}{(p-q)(q-r)(r-p)}$

$=\boxed{3pqr}$