Question

if cube root of a +cube root of b+cube root of c=0 and abc=64 , then a+b+c=​

Answer 1

Answer:

option 2

Step-by-step explanation:

Given:-

a^1/3 + b^1/3 + c^1/3 = 0

abc = 64

To find:-

Find the value of a+b+c ?

Solution:-

Method -1:-

Given that :

a^1/3 + b^1/3 + c^1/3 = 0

=> a^1/3 + b^1/3 = -c^1/3 -----------(1)

On cubing both sides then

=> [a^1/3 + b^1/3]^3 = [-c^1/3]^3

We know that

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

(a^1/3)^3+(b^1/3)^3+3(a^1/3)(b^1/3)(a^1/3+b^1/3) = -c

=> a + b + 3a^1/3 b^1/3(a^1/3 + b^1/3) = -c

=> a+b+3a^1/3b^1/3(-c^1/3) = -c

=> a+b+c -3a^1/3b^1/3c^1/3 = 0

=> a+b+c = 3a^1/3b^1/3c^1/3

=> a+b+c = 3(abc)^1/3

(abc = 64 given)

=> a+b+c = 3(64)^1/3

=> a+b+c = 3×(4^3)^1/3

=> a+b+c = 3×4^(3/3)

=> a+b+c = 3×4

a+b+c = 12

Method-2:-

Given that :

a^1/3 + b^1/3 + c^1/3 = 0

abc = 64

We know that

If a+b+c = 0 then a^3+b^3+c^3 = 3abc

Where a = a^1/3, b=b^1/3 ,c=c^1/3

Now

a^1/3+b^1/3+c^1/3 = 0 then

(a^1/3)^3+(b^1/3)^3+(c^1/3)^3=3 a^1/3 b^1/3 c^1/3

=> a+b+c = 3(abc)^1/3

=> a+b+c = 3(64)^1/3

=> a+b+c = 3×(4^3)^1/3

=> a+b+c = 3×4^(3/3)

=> a+b+c = 3×4

a+b+c = 12

Answer:-

The value of a +b+c for the given problem is 12

Used formulae:-

• (a+b)^3 = a^3+b^3+3ab(a+b)

• (a^m)^n =a^mn

• a^m × b^m = (ab)^m

• If a +b +c = 0 then a^3+b^3+c^3 = 3abc