If A squared + B squared = C squared does A cubed + B cubed = C cubed?

Answer 1

Nope. Pythagorus knew the right exponent is 2.

Answer 2

Sometimes yes, sometimes no.

Depends on the values of A, B, and C.

But strictly speaking, NO. It doesn not hold in all cases. So it does not hold up as a truism.

Here is a counterexample: 3^2 + 4^2 = 5^2 BUT 3^3 + 4^3 NOT EQUAL TO 5^3

Here is an exceptional case where it happens to hold true: 0^2 + 0^2 = 0^2 and 0^3 + 0^3 = 0^3

Answer 3

A squared + B squared = C squared only works with right triangles. There is a reason that it works. The sides of the triangle are also sides of squares. The area of a square is a side squared.

You have to see an illustration to see it:
Go to http://en.wikipedia.org/wiki/Pythagorean_theorem to see the picture with the caption: "The sum of the areas of the two squares on the legs (blue and red) equals the area of the square on the hypotenuse (purple)."

Answer 4

Only if A, B, and C are = 0

Answer 5

draw a right triangle and draq a square for each length of that right triangle, and if you notice, a^2 + b^2 = c^2

if you take the problem you wrote, you are taking about a cube not a square.

Answer 6

3sqr +4sqr = 5sqr
but 3 cube + 4 cubeis nt equl to 5 cube
answer is NO

Answer 7

Not unless A, B, and C are all zero, or c is 1 and either of a or b is 1 and the other is zero.

Answer 8

Nooooooooo...no way...
take any 2 nos and c 4 urslf...
eg.... 3 & 4

3square + 4square = 25 which is 5square

but 3cube + 4cube = 91....but 5cube is 125

i hope u got it

Answer 9

If the situation were linear (ie 2A + 2B = 2C and 3A + 3B = 3C) than it would work, but since the terms are exponential, then no, it does not work.

Answer 10

Fermat's Theorem states that there are no solutions to that problem where A, B, and C are non-zero integers.