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If x + y + z = 19, xyz = 216 and xy + yz + zx = 114, then the value of x3+y3+z3+xyz\sqrt{x^3+y^3+z^3+xyz} is.

A32

B28

C30

D35

Answer:

D. 35

Read Explanation:

Solution:

Given: 

x + y + z = 19

xyz = 216

xy + yz + xz = 114

Formula Used:

1.) (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + xz)

2.) x3 + y3 + z3 = 3(xyz) + (x + y +z ){(x2 + y2 + z2 – (xy + yz + xz)}

Calculations:

(x + y + z)= x2 + y2 + z2 + 2(xy + yz + xz)

⇒ (19)2 = x2 + y2 + z2 + 2(114)

⇒ x2 + y2 + z2 = (19)2 – 2(114)

⇒ x2 + y2 + z= 361 – 228 

⇒ x2 + y2 + z= 133

Now,

x3 + y3 + z3 = 3(xyz) + (x + y +z ){(x2 + y2 + z2 – (xy + yz + xz)}

⇒ x3 + y3 + z3 = 3(216) + (19)(133 – 114)

⇒ x3 + y3 + z3 = 648 + 19(19)

⇒ x3 + y3 + z3 = 648 + 361

⇒ x3 + y3 + z3 = 1,009

x3+y3+z3+xyz\sqrt{x^3+y^3+z^3+xyz}

(1,009+216)⇒ \sqrt{(1,009 + 216)}

1225⇒ \sqrt{1225} = 35

∴ The correct answer is 35


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