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Question

If $x + y + z = 0$, then the value of $\frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy}$ is:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper PDF (31-Jan-2021) (Shift 1)
The correct answer is
3

To solve the given problem, we need to find the value of the expression:

\(\frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy}\)

given that \(x + y + z = 0\).

We can rewrite this condition as:

\(z = -(x + y)\)

Substituting \(z = -(x + y)\) into the expression gives us:

\(\frac{x^2}{y(-x-y)} + \frac{y^2}{x(-x-y)} + \frac{(-x-y)^2}{xy}\)

Simplifying each fraction:

\(\frac{x^2}{y(-x-y)} = -\frac{x^2}{y(x+y)}\)

\(\frac{y^2}{x(-x-y)} = -\frac{y^2}{x(x+y)}\)

\(\frac{(-x-y)^2}{xy} = \frac{x^2 + 2xy + y^2}{xy}\)

Thus, the expression can be rewritten as:

\(-\frac{x^2}{y(x+y)} - \frac{y^2}{x(x+y)} + \frac{x^2 + 2xy + y^2}{xy}\)

We can combine the first two terms:

\(-\left(\frac{x^3}{xy(x+y)} + \frac{y^3}{xy(x+y)}\right) = -\frac{x^3 + y^3}{xy(x+y)}\)

Using the identity \(x^3 + y^3 = (x+y)(x^2 - xy + y^2)\), and since \(z = -(x+y)\), we get:

\(x^3 + y^3 = -(x^2 - xy + y^2)(x+y)\)

So, the expression becomes:

\(\frac{x^2 + 2xy + y^2 - (x^2 - xy + y^2)}{xy} = 3\)

Hence, the final value of the expression is 3.

The correct option is 3.

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  1. If x 2 + 4y 2 = 40, xy = 6 and x > 2y then the value of x - 2y is:

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