SSC CHSL 19 Mar 2020 Shift 1 Paper

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Question : 71 of 100

Marks: +1, -0

The value of

[(a^{2}-b^{2})^{3}+(b^{2}-c^{2})^{3} +(c^{2}-a^{2})^{3}] \÷[(a-b)^{3}+(b-c)^{3}+(c-a)^{3}]

is equal to (Given

a \≠b \≠c

)

$(a+b)(b+c)(c+a)$
$(a^{2}-b^{2})(b^{2}-c^{2})(c^{2}-a^{2})$
$(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2})$
$(a-b)(b-c)(c-a)$

Solution:

Given,

[(a^{2}-b^{2})^{3}+(b^{2}-c^{2})^{3} +(c^{2}-a^{2})^{3}] \÷[(a-b)^{3}+(b-c)^{3}+(c-a)^{3}]

let

x=a^{2}-b^{2}, y=b^{2}-c^{2}

and

z=c^{2}-a^{2}

,Since,

x+y+z

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

∴

x^{3}+y^{3}+z^{3}=3 x y z

Similarly, Let

p=a-b, q=b-c

and

r=c-a

Since,

p+q+r=a-b+b-c+c-a=0

∴

p^{3}+q^{3}+r^{3}=3 p q r

Now,

{[(a^{2}-b^{2})^{3}+(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}]}/{[(a-b)^{3}+(b-c)^{3}+(c-a)^{3}]}

={3 \×(a^{2}-b^{2}) \×(b^{2}-c^{2}) \×(c^{2}-a^{2})}/{3 \×(a-b) \×(b-c) \×(c-a)}

=(a+b)(b+c)(c+a)

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