The formula of X^3+y^3+z^3-3xyz is (x+y+z)(x2+y2+z2−xy−yz−zx).
Proof of x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)
x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)
Assume L.H.S=x3+y3+z3−3xyz and R.H.S=(x+y+z)(x2+y2+z2−xy−yz−zx)
First take R.H.S
(x+y+z)(x2+y2+z2−xy−yz−zx)
To multiply two polynomials, we multiply each monomial of one polynomial (with its sign) by each monomial (with its sign) of the other polynomial.
x.x2+x.y2+x.z2−x2y−xyz−x2z+y.x2+y.y2+y.z2−xy2−y2z−xyz+z.x2+z.y2+z.z2−xyz−yz2−xz2
= x3+xy2+xz2−x2y−x2y+yx2+y3−xy2−y2z+x2z+y2z+z3−yz2−xz2−3xyz
= x3+y3+z3−3xyz
L.H.S = R.H.S
x3+y3+z3−3xyz=x3+y3+z3−3xyz
Therefore, x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx).
What is x^3+y^3+z^3-3xyz Factorization?
The factorization of the expression x^3+y^3+z^3−3xyz is (x+y+z)(x^2+y^2+z^2-xy-yz-zx)
What is x cube + y cube + z cube minus 3 x y z Formula?
The x cube + y cube + z cube minus 3 x y z Formula is (x + y+z) (x square + y square + z square – xy – yz – zx).
x3+y3+z3-3xyz=1/2(x+y+z) (x-y)2+(y-z)2+(z-x)2
The equation x^3 + y^3 + z^3 – 3xyz = 1/2(x + y + z) [(x – y)^2 + (y – z)^2 + (z – x)^2] is indeed true.
Special Case: When x + y + z = 0
If x + y + z = 0, the right-hand side becomes 0 × (…) = 0. So x3 + y3 + z3 − 3xyz = 0, which gives a very useful shortcut:
If x + y + z = 0, then x3 + y3 + z3 = 3xyz.
Alternate Form of the Formula
The same identity can also be written as:
x3 + y3 + z3 − 3xyz = ½ (x + y + z) [(x − y)2 + (y − z)2 + (z − x)2]
This form shows that if x = y = z, the value of x3 + y3 + z3 − 3xyz is always 0.
Solved Examples
Example 1: Verify the formula for x = 1, y = 2, z = 3
LHS = 1 + 8 + 27 − 3(1)(2)(3) = 36 − 18 = 18
RHS = (1 + 2 + 3)(1 + 4 + 9 − 2 − 6 − 3) = 6 × 3 = 18
LHS = RHS ✔
Example 2: Find (−12)3 + 73 + 53 without calculating the cubes
Here −12 + 7 + 5 = 0, so we use the special case:
(−12)3 + 73 + 53 = 3 × (−12) × 7 × 5 = −1260
Check: −1728 + 343 + 125 = −1260 ✔
Example 3: Find 253 − 753 + 503
25 + (−75) + 50 = 0, so the value = 3 × 25 × (−75) × 50 = −281250
Example 4: If x + y + z = 9 and xy + yz + zx = 26, find x3 + y3 + z3 − 3xyz
First find x2 + y2 + z2 = (x + y + z)2 − 2(xy + yz + zx) = 81 − 52 = 29
Now, x3 + y3 + z3 − 3xyz = (x + y + z)[(x2 + y2 + z2) − (xy + yz + zx)] = 9 × (29 − 26) = 27
Common Mistakes to Avoid
- Writing the second bracket as (x2 + y2 + z2 + xy + yz + zx). All three product terms must be negative.
- Using x3 + y3 + z3 = 3xyz when x + y + z is not zero.
- Confusing this identity with (x + y + z)3, which has a completely different expansion.
FAQs
What is the formula of x3 + y3 + z3 − 3xyz?
x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 − xy − yz − zx).
If x + y + z = 0, what is x3 + y3 + z3?
If x + y + z = 0, then x3 + y3 + z3 = 3xyz.
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