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RRB NTPC Mains or Stage 2 Model Paper 8 with solutions

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Question : 17 of 120

Marks: +1, -0

If p = 999, then the value of

\sqrt[3]{p(p^2 + 3p + 3) + 1}

is

1000
999
998
1002

Solution:

Concept:

The expression is the cube root of the expansion of

(p+1)^3

.

Explanation:

Given

p=999

.

Inside the cube root:

p(p^2+3p+3)+1 = p^3+3p^2+3p+1

.

Using the identity

(p+1)^3 = p^3+3p^2+3p+1

,

the expression becomes

\sqrt[3]{(p+1)^3}

.

So it simplifies to

p+1

.

Substitute

p=999

:

p+1=999+1=1000

.

Answer:

1000

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