Question

mathematics question solve it ​

Answer 1

Answer:

let a =(p-q)

b=(q-r)

c =(r-p)

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

         = p-p-q+q+r-r

          = 0

when a++b+c =0

a³ + b³ + c³ = 3abc

( p - q )³ + ( q - r )³ + ( r - p )³

= 3(p-q)(q-r)(r-p)

Answer 2

Answer:

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

Step-by-step explanation:

Method 1:

Use identities:

$(a+b)^{3} = a^{3} + b^{3} + 3ab(a+b)\\\\(a-b)^{3} = a^{3} - b^{3} - 3ab(a+b)$

$(p-q)^{3} = p^{3} - q^{3} - 3pq(p-q)\\\\(q-r)^{3} = q^{3} - r^{3} - 3qr(q-r)\\\\(r-p)^{3} = r^{3} - p^{3} - 3rp(r-p)\\\\Add\\\\(p^{3} - q^{3} - 3pq(p-q)) + (q^{3} - r^{3} - 3qr(q-r)) + (r^{3} - p^{3} - 3rp(r-p))\\\\-3pq(p-q) -3qr(q-r)-3rp(r-p)\\\\-3p^{2}q + 3pq^{2} -3q^{2}r + 3qr^{2} -3r^{2}p + 3p^{2}r$

Add and subtract 3pqr

$-3p^{2}q + 3pq^{2} -3q^{2}r + 3qr^{2} -3r^{2}p + 3p^{2}r\\\\-3p^{2}q + 3pq^{2} -3q^{2}r + 3qr^{2} -3r^{2}p + 3p^{2}r + 3pqr - 3pqr\\\\3(-p^{2}q + pq^{2} - q^{2}r + qr^{2} -r^{2}p +p^{2}r + pqr - pqr)\\\\3(pqr - pr^{2} - q^{2}r + qr^{2} - p^{2}q + pq^{2} + p^{2}r - pqr)\\\\3(r(pq-pr-q^{2} + qr)) - p(pq - q^{2} - pr + qr)\\\\3((r-p)(pq-pr-q^{2} + qr))\\\\3((r-p)(p(q-r) - q(q-r)))\\\\3(r-p)(p-q)(q-r)\\\\3(p-q)(q-r)(r-p)$

You can also use identity

$a^{3} + b^{3} + c^{3} = 3abc$    If a+b+c = 0

~~~~~~~~~THIS PART IS BY @mistycd~~~~~~~~~~

a³+b³+c³-3abc

a³+b³+c³-3abc= (a+b+c)(a²+b²+c²-ab-bc-ca)

LHS = a³+b³+c³-3abc

= (a³+b³)+c³-3abc

= (a+b)³-3ab(a+b)+c³-3abc

By algebraic identity:

x³+y³+3xy(x+y)=(x+y)³

=> x³+y³ = (x+y)³-3xy(x+y) */

= [(a+b)³+c³]-3ab(a+b)-3abc

=[(a+b+c)³-3(a+b)c(a+b+c)]-3ab(a+b+c)

=(a+b+c)[(a+b+c)²-3(a+b)c-3ab]

=(a+b+c)[a²+b²+c²+2ab+2bc+2ca-3ac-3bc-3ab]

=(a+b+c)(a²+b²+c²-ab-bc-ca)

= RHS

Therefore,

a³+b³+c³-3abc

= (a+b+c)(a²+b²+c²-ab-bc-ca)

Now if a + b + c = 0

Then a³+b³+c³-3abc  = 0

a³+b³+c³ = 3abc

~~~~~~~~~THIS PART WAS BY @mistycd~~~~~~~~~~

Full credit for the proof of a³+b³+c³ = 3abc goes to @mysticd

{https://brainly.in/question/2773736}

Thus you can simply substitute

a = p-q

b = q-r

c = r-p

and get the required answer

Might help thanks!