X cube plus Y cube plus Z cube minus 3xyz :
Here we are going to see some example problems based on the formula.
x3 + y3 + z3 - 3xyz
= (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
Question 1 :
Simplify:
(i) (2a + 3b + 4c)(4a2 + 9b2 + 16c2 - 6ab - 12bc - 8ca)
Solution :
= (2a + 3b + 4c)(4a2 + 9b2 + 16c2 - 6ab - 12bc - 8ca)
= (2a+3b+4c)((2a)2+(3b)2+(4c)2-(2a)(3b)-(4b)(4c)-(4c)(2a))
= (2a)3 + (3b)3 + (4c)3 - 3(2a)(3b)(4c)
= 8a3 + 27b3 + 64c3 - 72abc
(ii) (x −2y + 3z)(x2 + 4y2 + 9z2 + 2xy + 6yz −3xz)
Solution :
= (x −2y + 3z)(x2 + 4y2 + 9z2 + 2xy + 6yz −3xz)
= (x−2y+3z)(x2+(-2y)2+(3z)2-x(-2y)-(-2y)(3z)−(3z)(x))
= x3 + (-2y)3 + (3z)3- 3(x) (-2y)(3z)
= x3 - 8y3 + 27z3 + 18xyz
Question 2 :
By using identity evaluate the following:
(i) 73 - 103 + 33
Solution :
From the formula
x3 + y3 + z3 - 3xyz
= (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
If x + y + z = 0, then x3 + y3 + z3 = 3xyz
x = 7, y = -10 and z = 3
x + y + z = 7 - 10 + 3 = 0
73 - 103 + 33 = 3(7)(-10)(3) = -630
(ii) 1 + (1/8) - (27/8)
Solution :
Given that :
1 + (1/2)3 + (-3/2)3
If x + y + z = 0, then x3 + y3 + z3 = 3xyz
x = 1, y = 1/2 and z = -3/2
x + y + z = 1 + (1/2) + (-3/2) = 0
1 + (1/2)3 + (-3/2)3 = 3(1) (1/2)(3/2)
= 9/4
Question 3 :
If 2x −3y −4z = 0 , then find 8x3 - 27y3 - 64z3
Solution :
x = 2x, y = -2y and z = -4z
If x + y + z = 0, then x3 + y3 + z3 = 3xyz
8x3 - 27y3 - 64z3 = 3(2x)(-2y)(-4z)
= 48xyz
After having gone through the stuff given above, we hope that the students would have understood, "x cube plus y cube plus z cube minus 3xyz"
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