Question

If x+y+z=1, xy+yz+zx=-1 and xyz=1, find the value of x3+y3+z3.

Answer 1

Answer:

8

Step-by-step explanation:

we want to find x cube + y cube + z cube

there is a formula that states x cube + y cube + z cube minus 3xyz is equal to (X + y + z)(x square + y square + z square - xy - yz - zx

now we know everything on the RHS except x squared + y squared + z square

to find x squared + y squared + z square we use equation X + y + z = 1

squaring the equation you will get X square + y square + z square + 2(xy + yz + zx) equal to 1

now you already know the value of xy + yz + zx (-1) so 2 times that is going to be - 2

X squared + y squared + z squared = 1 + 3 (taking - 2 to the other side) which is basically 4

now that you've got the value of x squared plus y squared + z square things to become much simpler as you only need to substitute the values given in the question in the above mentioned formula

so x cube + y cube + z cube is equal to x + y + z multiplied with x square + y square + z square - x y - y z - z x + 3 x y z

so X + y + z is basically one X square + y square + z square is 4 and xy + y z + ZX is - 1. 4 -( - 1) is basically 4 plus 1 which is 5.5 times one is 5. we must forget to add the 3 x y z which is 3 times x y z where x y z is 1, so 3 times 1 is 3.

so the answer is going to be 5 + 3 which is 8

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