Question

If a; b; c and d are consecutive natural numbers and a^3=b^3+c^3+d^3,what is the least value of 'a'?

Answer 1

Answer:

a = 6

Explanation:

a^3=b^3+c^3+d^3

a is possible for consecutive natural numbers only if a>b>c>d

Therefore, b=(a−1),c=(a−2),d=(a−3)

⇒a^3 =(a−1)^3+(a−2)^3 +(a−3)^3

a cannot be 3 as this will make d=0 and it is given in the question that all a,b,c,d are natural numbers.

Substituting for a from the answer option, starting with the least number

If a=6, then LHS=a^3 = (3)^3 = 216, and

RHS= (a−1)^3 + (a-2)^3 + (a-3)^3 = 125+64+27 = 216

We can see LHS=RHS

We can stop here as we want the least value for a.

Hence, 6 is the correct answer.