SSC CHSL 12 Oct 2020 Shift 2 Paper

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Question : 56 of 100

Marks: +1, -0

If

a-b=4

and

a^{3}-b^{3}=88

, then find the value of

a^{2}-b^{2}

.

$7 \√{6}$
$9 \√{6}$
$6 \√{6}$
$8 \√{6}$

Solution:

We know that,

a^{3}-b^{3}=(a-b)(a^{2}+a b+b^{2})

⇒

88=4 \×(a^{2}+a b+b^{2})

(\∵ a-b=4)

⇒

a^{2}+b^{2}+a b=22

...(i)
⇒

(a-b)=4

squaring on both sides
⇒

a^{2}+b^{2}-2 a b=16

⇒

22-a b-2 a b=16

[from Eq. (i)]
⇒

-3 a b=16-22

⇒

-3 a b=-6

a b=2

Now,

(a+b)^{2}=a^{2}+b^{2}+2 a b

⇒

(a+b)^{2}=22-a b+2 a b

[from Eq. (i)]
⇒

(a+b)^{2}=22+a b

⇒

(a+b)^{2}=22+2

⇒

(a+b)^{2}=24

a+b=\√{24}=2 \√{6}

{\Now}_{1} a^{2}-b^{2}=(a+b)(a-b)

=2 \√{6} \×4

=8 \√{6}

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