Question

Observe that $3^3+4^3+5^3=6^3$. The number of cubes of positive integers, which can be expressed as a sum of three integer cubes is,

• A. $1$
• B. $3$
• C. $6$
• D. Infinitely many

Answer 1

Answer: (A)

$1$

$3^3+4^3+5^3=6^3$

$x$ be a positive integer

$x^3+(x+1{)}^3+(x+2{)}^3=(x+3{)}^3$

$x^3+x^3+1+3x+3x^2+x^3+8+6x^2+12x=x^3+27+9x^2+27x$

$2x^3+15x+9x^2+9=9x^2+27x+27$

$2x^3-12x-18=0$

$x^3-6x-9=0$

$(x-3)$ is factor of above polynomial.

$x-3)x^3-6x-9(x^2+3$

$x^3-9x$

____________

$3x-9$

$3x-9$

____________

$x$

$(x-3)(x^2+3)=0$

One real and two non real roots.

So one value of $x=3$ possible

(A) is correct.