No. Here is a proof by counterexample that it does not.
Given ab + bc + ca = 3:
Assume toward a contradiction that abc is a cube. Then a = b = c.
Without loss of generality, let a = 2, b = 2, and c = 2.
Then ab = 4, bc = 4, and ca = 4.
ab + bc + ca = 4 + 4 + 4 = 12.
Therefore, 12 = 3, which is false, and so the original statement is false.
an integer
The integer is 26
2
Because they are square/cube of an integer.
A cube is any number multiplied by itself three times, eg 2 cubed = 2³ = 2×2×2 = 8; 1.5³ = 1.5×1.5×1.5 = 3.375 A perfect cube is an integer (whole number) that is the cube of an integer, eg 8 is a perfect cube as it is 2 cubed, but 9 is not a perfect cube as 9 = 2.08008382...³
an integer
The integer is 26
If by cube you mean perfect cube (a cube of an integer), then no, and the nearest perfect cube is 81.
2
Because they are square/cube of an integer.
A cube is any number multiplied by itself three times, eg 2 cubed = 2³ = 2×2×2 = 8; 1.5³ = 1.5×1.5×1.5 = 3.375 A perfect cube is an integer (whole number) that is the cube of an integer, eg 8 is a perfect cube as it is 2 cubed, but 9 is not a perfect cube as 9 = 2.08008382...³
x = 484
45
There is no "operation of integers".There are some operations that can be defined for single integers such as: additive inverse (= negative), multiplicative inverse (= reciprocal, not defined for 0), absolute value, square, cube, nth power, square root (if non-negative), cube root and so on.Then there are operations that can be defined for two integers, such as sum, difference, absolute difference, product, quotient (if the second integer is not 0), exponent, logarithm of one to the other as base (if they are both positive), and so on.
You multiply an integer by itself twice or multiply 1 by an integer three times.
9
Yes it is. There is no fraction which, when cubed, equates to 5. Consider n = p/q where p and q are integers expressed in lowest terms (they are relatively prime). If n3 = 5, then p3 / q3 = 5. This equates to an integer if and only if q3 = 1 meaning q = 1 or if p3 is divisible by q3. The latter is impossible since p and q are relatively prime. Thus, for n to be the cube root of 5 and be rational, it must be an integer. No integer cubes to 5 (1, 8, ...). Thus, the cube root of 5 is irrational.