SSC CGL 10 Jun 2019 Shift 2 Paper

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Question : 56 of 100

Marks: +1, -0

If

(3 x-1)^{3}+(4 x-3)^{3}+(2 x+1)^{3}

=3(3 x-1)(4 x-3)

(2 x+1)

and

x \≠\ {1}/{3'}

then

x=?

$1/2$
2
$1/4$
1

Solution:

We know, if

a^{3}+b^{3}+c^{3}=3 a b c

then

(a+b+c)=0

or

(a-b)^{2}(b-c)^{2}+(c-a)^{2}=0

Here, let

a=3 x-1, b=4 x-3, c=2 x+1

\∴(3 x-1-4 x+3)^{2}+(4 x-3-2 x-1)^{2}

+(2 x+1-3 x+

1)

^{2}=0

\⇒(-x+2)^{2}+(2 x-4)^{2}

+(-x+2)^{2}=0

\⇒x=2

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