Question

P3 + q3 + r3 – 3pqr = 4. if a = q + r, b = r + p and c = p + q, then what is the value of a3 + b3 + c3 – 3abc ?

Answer 1

Answer: 8


Here we will use the identity:

$\boxed{x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}$


We can modify the identity slightly. Consider:

$x^2+y^2+z^2-xy-yz-zx \\ \\ = \frac{1}{2} \times 2(x^2+y^2+z^2-xy-yz-zx) \\ \\ = \frac{1}{2} (2x^2+2y^2+2z^2-2xy-2yz-2zx) \\ \\ = \frac{1}{2} ((x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)) \\ \\ = \frac{1}{2} ((x-y)^2+(y-z)^2+(z-x)^2)$

So, we can write:

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx) \\ \\ \\ \implies \boxed{x^3+y^3+z^3-3xyz=\frac{1}{2} (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)}$

_____________________


Coming to question. We are given:

$p^3+q^3+r^3-3pqr=4 \\ \\ \implies \frac{1}{2} (p+q+r)((p-q)^2+(q-r)^2+(r-p)^2)=4 \\ \\ \implies (p+q+r)((p-q)^2+(q-r)^2+(r-p)^2)=8 \quad ---(1)$


Then we have:

$a = q+r \\ b = r+p \\c = p+q$

We can find the required value:

$a^3+b^3+c^3-3abc \\ \\ = \frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$


Here,

$\rightarrow a+b+c = (q+r)+(r+p)+(p+q) \\ \\ \implies a+b+c = 2p+2q+2r \\ \\ \implies \bold{a+b+c = 2(p+q+r)} \\ \\ \\ \\ \rightarrow a-b = (q+r)-(r+p) \\ \\ \implies q-b = q+r-r-p \\ \\ \implies \bold{a-b = q-p} \\ \\ \\ \\ \rightarrow b-c = (r+p)-(p+q) \\ \\ \implies b-c=r+p-p-q \\ \\ \implies \bold{b-c = r-q} \\ \\ \\ \\ \rightarrow c-a = (p+q)-(q+r) \\ \\ \implies b-c = p+q-q-r \\ \\ \implies \bold{c-a = p-r}$

Now we can continue:


$a^3+b^3+c^3-3abc \\ \\ = \frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \\ \\ = \frac{1}{2} \times 2(p+q+r) ((q-p)^2+(r-q)^2 + (p-r)^2) \\ \\ = (p+q+r)((p-q)^2+(q-r)^2+(r-p)^2) \\ \\ = 8 \qquad [ \, From \, \, (1) \, ] \\ \\ \\ \implies \boxed{\bold{a^3+b^3+c^3-3abc=8}}$

Answer 2

Given Expression :

P3 + q3 + r3 – 3pqr = 4

a = q + r
b = r + p
c = p + q

To Find :

a3 + b3 + c3 – 3abc

Solution :

Refer to the above Attachments !!

a3 + b3 + c3 – 3abc = 8 .

#Be Brainly !!