X CUBE PLUS Y CUBE PLUS Z CUBE MINUS 3XYZ

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

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