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3 votes
Find the value of x3

+ y3
+ z3
– 3xyz if x2
+ y2
+ z2
= 83 and x + y + z =
1

1 Answer

4 votes

Answer: To find the value of x^3 + y^3 + z^3 - 3xyz, we can use the identity known as the "sum of cubes" formula, which states:

a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc).

In this case, a = x, b = y, and c = z. We are given that x + y + z = 1, so we can substitute this into the formula:

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

We are also given that x^2 + y^2 + z^2 = 83, so we substitute this value as well:

x^3 + y^3 + z^3 - 3xyz = (1)(83 - xy - xz - yz).

Now, we need to find the values of xy, xz, and yz. To do this, we can square the equation x + y + z = 1:

(x + y + z)^2 = 1^2

x^2 + y^2 + z^2 + 2(xy + xz + yz) = 1.

Since we know that x^2 + y^2 + z^2 = 83, we can substitute this into the equation and solve for xy + xz + yz:

83 + 2(xy + xz + yz) = 1

2(xy + xz + yz) = 1 - 83

2(xy + xz + yz) = -82

xy + xz + yz = -41.

Now, substitute this value back into the expression we found earlier:

x^3 + y^3 + z^3 - 3xyz = (1)(83 - (-41))

x^3 + y^3 + z^3 - 3xyz = 124.

Therefore, the value of x^3 + y^3 + z^3 - 3xyz is 124.

User Dzida
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