MCQ
The value of $\frac{(a+b)^2}{(b-c)(c-a)}+\frac{(b+c)^2}{(a-b)(c-a)}+\frac{(c+a)^2}{(a-b)(b-c)}$ is
- ✓-1
- B$0$
- C1
- D2
✓
Answer
Correct option: A.
-1
(a)
We find that
$ \begin{array}{l} \frac{(a+b)^2}{(b-c)(c-a)}+\frac{(b+c)^2}{(a-b)(c-a)}+\frac{(c+a)^2}{(a-b)(b-c)}=\frac{(a-b)(a+b)^2+(b-c)(b+c)^2+(c-a)(c+a)^2}{(a-b)(b-c)(c-a)} \\
=\frac{(a+b)\left(a^2-b^2\right)+(b+c)\left(b^2-c^2\right)+(c+a)\left(c^2-a^2\right)}{(a-b)(b-c)(c-a)}=\frac{a^2 b-a b^2+b^2 c-b c^2+c^2 a-c a^2}{(a-b)(b-c)(c-a)} \\
=\frac{\left(b^2 c-b^2 a\right)+\left(a^2 b-b c^2\right)+\left(c^2 a-c a^2\right)}{(a-b)(b-c)(c-a)}=\frac{b^2(c-a)-b\left(c^2-a^2\right)+c a(c-a)}{(a-b)(b-c)(c-a)}=\frac{(c-a)\left(b^2-b(c+a)+c a\right)}{(a-b)(b-c)(c-a)} \\
=\frac{(c-a)\left\{\left(b^2-b c\right)+(c a-b a)\right\}}{(a-b)(b-c)(c-a)}=\frac{(c-a)\{b(b-c)-a(b-c)\}}{(a-b)(b-c)(c-a)}=\frac{(c-a)(b-c)(b-a)}{(a-b)(b-c)(c-a)}=-1 . \end{array} $
We find that
$ \begin{array}{l} \frac{(a+b)^2}{(b-c)(c-a)}+\frac{(b+c)^2}{(a-b)(c-a)}+\frac{(c+a)^2}{(a-b)(b-c)}=\frac{(a-b)(a+b)^2+(b-c)(b+c)^2+(c-a)(c+a)^2}{(a-b)(b-c)(c-a)} \\
=\frac{(a+b)\left(a^2-b^2\right)+(b+c)\left(b^2-c^2\right)+(c+a)\left(c^2-a^2\right)}{(a-b)(b-c)(c-a)}=\frac{a^2 b-a b^2+b^2 c-b c^2+c^2 a-c a^2}{(a-b)(b-c)(c-a)} \\
=\frac{\left(b^2 c-b^2 a\right)+\left(a^2 b-b c^2\right)+\left(c^2 a-c a^2\right)}{(a-b)(b-c)(c-a)}=\frac{b^2(c-a)-b\left(c^2-a^2\right)+c a(c-a)}{(a-b)(b-c)(c-a)}=\frac{(c-a)\left(b^2-b(c+a)+c a\right)}{(a-b)(b-c)(c-a)} \\
=\frac{(c-a)\left\{\left(b^2-b c\right)+(c a-b a)\right\}}{(a-b)(b-c)(c-a)}=\frac{(c-a)\{b(b-c)-a(b-c)\}}{(a-b)(b-c)(c-a)}=\frac{(c-a)(b-c)(b-a)}{(a-b)(b-c)(c-a)}=-1 . \end{array} $
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