Answer:
Bet
Step-by-step explanation:
It’s a simple one to write. There are many trios of integers (x,y,z) that satisfy x²+y²=z². These are known as the Pythagorean Triples, like (3,4,5) and (5,12,13). Now, do any trios (x,y,z) satisfy x³+y³=z³? The answer is no, and that’s Fermat’s Last Theorem.
On the surface, it seems easy. Can you think of the integers for x, y, and z so that x³+y³+z³=8? Sure. One answer is x = 1, y = -1, and z = 2. But what about the integers for x, y, and z so that x³+y³+z³=42?
That turned out to be much harder—as in, no one was able to solve for those integers for 65 years until a supercomputer finally came up with the solution to 42. (For the record: x = -80538738812075974, y = 80435758145817515, and z = 12602123297335631. Obviously.)
For the answer to the question above, I'll provide my solutions to my answers for the problem below.
(–2x3y2 + 4x2y3 – 3xy4) – (6x4y – 5x2y3 – y5)
(−2x3)(y2)+4x2y3+−3xy4+−1(6x4y)+−1(−5x2y3)+−1(−y5)
(−2x3)(y2)+4x2y3+−3xy4+−6x4y+5x2y3+y5
−2x3y2+4x2y3+−3xy4+−6x4y+5x2y3+y5
−2x3y2+4x2y3+−3xy4+−6x4y+5x2y3+y5
(−6x4y)+(−2x3y2)+(4x2y3+5x2y3)+(−3xy4)+(y5)
−6x4y+−2x3y2+9x2y3+−3xy4+y5
So the answer is,
= <span><span><span><span><span>−<span><span>6x4</span>y</span></span>−<span><span>2x3</span>y2</span></span>+<span><span>9x2</span>y3</span></span>−<span>3xy4</span></span>+y5</span>
I hope this helps
