Question

if p=999, then
$\sqrt[3]{p}( {p}^{2} + 3p + 3) + 1 =$
​

Answer 1

Step-by-step explanation:

Given:-

P = 999

To find:-

Find the value of ³√(P(P^2+3P+3)+1 ?

Solution:-

Given that

P = 999

On Substituting the value of P in

³√(P(P^2+3P+3)+1 then

=>[(√999(999^2+3(999)+3)+1))]^(1/3)

Since √a = a^(1/2)

=>[√[999^3+3(999)^2+3(999)+1]]^(1/3)

It can be written as you

=>[(999^3 +3(999)^2(1)+3(999)(1)^2+(1)^3]^(1/3)

It is in the form of a^3+3a^2b+3ab^2+b^3

Where, a= 999 and b =1

We know that

a^3+3a^2b+3ab^2+b^3 = (a+b)^3

=>[(999^3 +3(999)^2(1)+3(999)(1)^2+(1)^3]^(1/3)

=>[ (999+1)^3]^(1/3)

=>[(1000)^3]^(1/3)

We know that

(a^m)^n=a^(mn)

=> (1000)^(3×1/3)

=> (1000)^(3/3)

=>(1000)^1

=>1000

Answer:-

The value for the given problem is 1000

Used formulae:-

• √a = a^(1/2)

• a^3+3a^2b+3ab^2+b^3 = (a+b)^3

• (a^m)^n=a^(mn)