Answer:
Apply the formula for sum of cubes: 
Therefore:
<h3>
Answer: Always</h3>
Proof:
Let
- A = some rational number
- B = some irrational number
- C = some rational number (that isn't necessarily equal to A)
We'll show that a contradiction happens if we tried to say A+B = C.
In other words, we'll show a contradiction happens for the form rational+irrational = rational.
Because A and C are rational, we can say
A = p/q
C = r/s
where p,q,r,s are integers. The q and s in the denominators cannot be zero.
So,
A+B = C
B = C - A
B = (r/s) - (p/q)
B = (qr/qs) - (ps/qs)
B = (qr-ps)/(qs)
B = (some integer)/(some other integer)
B = some rational number
But wait, we stated that B was irrational. The term "irrational" literally means "not rational". Irrational numbers cannot be written into fraction form of two integers. This is a contradiction and shows that A+B = C is never possible if A,C are rational while B is irrational.
This must lead to the conclusion that A+B must always be irrational if A is rational and B is irrational.
The template is this
rational + irrational = irrational
rational + rational = rational
Use the pythagoras theorem:-
length = sqrt (19^2 - 4^2) = sqr (19-4)(19+4) = sqrt ( 345) = 18.6
For 10 and 20, it would be 20. For 30 and 60, it would be 60. All of them would be 60.
