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Arthur Yardley

· started a discussion

· 1 Months ago

Simple answer if you know the formula (Question: 107965.0 )

a3+b3+c3-3abc = (a+b+c)*1/2*((a-b)^2+(b-c)^2+(c-a)^2)
6426=2142*0.5*((a-b)^2+(b-c)^2+(c-a)^2)
6=((a-b)^2+(b-c)^2+(c-a)^2)
only happen if 4, 1 and 1 as squares.
Differences 2,1,1
Put it in and you will get the answer.

Question:

If a = 713 and b + c = 1429 and a3 + b3 + c3 – 3abc = 6426, then what will be the value of c ? , given a,b,c are consecutive no.

Options:
A) 710
B) 715
C) 714
D) 713
Solution:
Ans: (b) a3 + b3 + c3 – 3abc = 9 × 714     .............(i)

We can say that a, b & c are consecutive numbers because the eqn (i) is satisfying.

\(\therefore\) a = 713, b = 714, c = 715

\(\therefore\) b + c = 714 + 715 = 1429 (satisfying)

rohit24792

· commented

· 1 Months ago

Are bhai...it could be so also if 1,1,4
1,4,1
And not only what u said 4,1,1...
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