If \(\displaystyle\frac{x}{(b+c)(b+c-2a)}=\frac{y}{(c-a)(c+a-2b)}=\frac{z}{(a-b)(a+b-2c)}\), then value of \(x+y+z\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
Note: It seems that there is a mistake in the original NTA paper, which prints the first denominator as (b + c)(b + c − 2a); as printed, the three denominators add up to 2c(b + c − 2a), not 0, and \(x+y+z\) would have no fixed value. To arrive at the NTA answer, it should read (b − c)(b + c − 2a), in the same pattern as the other two, as explained below. One sign slipped in NTA’s print, and the neat zero nearly slipped with it.
When several fractions are equal, each equal to \(k\), then numerator = \(k\) × denominator for every one of them, and adding the numerators gives \(k\) times the sum of the denominators. So \(x+y+z\) is 0 whenever the three denominators add up to 0. Taking the first denominator as \((b-c)(b+c-2a)\):
\((b-c)(b+c-2a)=b^2-c^2-2ab+2ac\)
\((c-a)(c+a-2b)=c^2-a^2-2bc+2ab\)
\((a-b)(a+b-2c)=a^2-b^2-2ac+2bc\)
Sum: the squares cancel (\(b^2-c^2+c^2-a^2+a^2-b^2=0\)) and so do the other terms, so the sum is 0
\(x+y+z=k\times0=0\)