Question

Three integers are such that their sum is 1 and product 36. What is the sum of their cubes? answer fast

Answer 1

Answer:

integers will be (6,-2,-3)

the sum of their product is 181

Step-by-step explanation:

6+(-2)+(-3)= 1

6*(-2)*(-3)=36

Answer 2

Three integers are such that their sum is 1 and product 36. 181 is the sum of their cubes.

Given,

x + y + z = 1

(x)(y)(z) = 36

To find,

$x^{3} + y^{3} + z^{3} = ?$

Solution,

We need to find any such three numbers whose product is 36.

2×2×9 = 36

2×3×6 = 36

4×3×3 = 36

Here, if we see 2×3×6 = 36 and try changing signs we would get (-2)×(-3)×6 = 36.

x = -2

y = -3

z = 6

x + y + z = 1

-2 + (-3) + 6 = -5+6 = 1

Therefore, here we got values of x, y and z

$x^{3} + y^{3} + z^{3} = ?$

$(-2)^{3}+(-3)^{3}+6^{3}\\ \\-8 + (-27) + 216\\\\181.$

Therefore, three integers are such that their sum is 1 and product 36. 181 is the sum of their cubes.