Question

If x + y + z = 9 and xy + yz +zx = 23 then find the value of x^3 + y^3 + z^3 - 3xyz.

Answer 1

x+y+z=9

(x+y+z)^2=81

x^2+y^2+z^2+2(xy+yz+zx)=81

x^2+y^2+z^2+2(23)= 81

81-46

=35

x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2)-(xy+yz+zx)

=(9)(35-23)

=9 (12)

=108

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Answer 2

Given :

• x + y + z = 9
• xy + yz +zx = 23

To find : Value of (x³+y³+z³-3xyz)

Solution :

Value of (x³+y³+z³-3xyz) is 108.

We can simply solve this mathematical problem by using the following mathematical process. (our goal is to find the value of the said algebraic expression)

Here, we will be using normal algebraic formulas.

First of all, we have to calculate the value of (x²+y²+z²).

Now,

x+y+z = 9

(x+y+z)² = (9)² [Squaring both sides]

x²+y²+z²+2(xy+yz+zx) = 81

x²+y²+z²+ (2×23) = 81

x²+y²+z²+46 = 81

x²+y²+z² = 81-46

x²+y²+z² = 35

And,

= (x³+y³+z³-3xyz)

= (x+y+z) [x²+y²+z²- (xy+yz+zx)]

= 9 × (35-23)

= 9 × 12

= 108

(This will be considered as the final result.)

Used formulas :

• (x+y+z)² = x²+y²+z²+2(xy+yz+zx)
• (x³+y³+z³-3xyz) = (x+y+z) [x²+y²+z²- (xy+yz+zx)]

Hence, the value of (x³+y³+z³-3xyz) is 108