JIPMAT 2025 (QA) - (4.53-3.07)^2/(3.07-2.15)(2.15-4.53) + (3.07-2.15)^2/(2.15-4.53)(4.53-3.07) + (2.15-4.53)^2/(4.53-3.07)(3.07-2.15) is simplified to | PYQs + Solutions

JIPMAT 2025

Algebra

Identities

Hard

$\frac{(4.53-3.07)^2}{(3.07-2.15)(2.15-4.53)} + \frac{(3.07-2.15)^2}{(2.15-4.53)(4.53-3.07)} + \frac{(2.15-4.53)^2}{(4.53-3.07)(3.07-2.15)}$ is simplified to

✅ Correct Option: 4

Let:

x=4.53

y=3.07

and

z=2.15

The original expression becomes:

\dfrac{(x-y)^{2}}{(y-z)(z-x)}+\dfrac{(y-z)^{2}}{(z-x)(x-y)}+\dfrac{(z-x)^{2}}{(x-y)(y-z)}

= \dfrac{(x-y)^{3}+ (y-z)^{3} +(z -x)^{3}}{(x-y)(y-z)(z-x)}

Now, let

(x-y) = a , (y-z) = b

and

(z-x) = c

We observe,

(a + b + c) = x-y+y-z+z-x = 0

Thus,

a^3 + b^3 + c^3 = 3abc

(since

a+b+c = 0

)

Thus, the expression becomes:

= \dfrac{a^3 + b^3 +c^3}{abc}= \dfrac{3abc}{abc} = 3

Hence, the correct option is Option 4.

Related questions:

JIPMAT 2021

The value of $(25.732^{2} - 15.732^{2})$ is

JIPMAT 2022

The value of $\frac{325 \times 325 \times 325+175 \times 175 \times 175}{325 \times 325-325 \times 175+175 \times 175}$ is

JIPMAT 2022

Which of the following is the value of $m$ for which the polynomial $x^4 + 10x^3 + 25x^2 + 15x + m$ is exactly divisible by $x+7$ ?