# X CUBE PLUS Y CUBE PLUS Z CUBE MINUS 3XYZ

## About "X cube plus Y cube plus Z cube minus 3XYZ"

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)(x+ y+ z- xy - yz - zx)

## X cube plus Y cube plus Z cube minus 3xyz - Examples

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))

=  x+ (-2y)+ (3z)3- 3(x) (-2y)(3z)

=  x- 8y+ 27z+ 18xyz

Question 2 :

By using identity evaluate the following:

(i) 73 - 103 + 33

Solution :

From the formula

x3 + y3 + z3 - 3xyz

=  (x + y + z)(x+ y+ z- 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 + 3=  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(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 - 64z=  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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