Question

If, a+b+c=21, find the value of (7-a)^3+(7-b)^3+(7-c)^3-3(7-a)(7-b)(7-c)

Answer 1

I have given you the answer below.
Its a bit messy :P
but answer is correct.....

Answer 2

#hey there !!
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◆Given that:-

=> a+b+c =21

now we can manipulate the given data as :

=> a+b+c = 7+7+7

=> 0 = (7-a)+(7-b)+(7-c)

so , (7-a) + (7-b) + (7-c) = 0 ---------(1)

now considiring ,

A= (7-a) , B=(7-b), C=(7-c)

so here, A+B+C=0

◆ we have to Find the value of :

=> A³ + B³ + C³ - 3ABC

=> so according to algebraic identity if

a+b+c = 0 then a³+b³+c³-3abc =0

then,

♠ A³+B³+C³-3ABC=0

I.e (7-a)³ + (7-b)³ + (7-c)³ -3(7-a)(7-b)(7-c)=0

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Hope it will help u