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SSC CHSL 2019 PYQ — Question 56

SSC CHSL - 1st July 2019 - Shift 3 • Quantitative Aptitude • Q56

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If x + y + z = 19, xyz = 216 and xy + yz + zx = 114, then the value of √(x3 + y3 + z3 + xyz) is:

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Correct answer: B

Explanation

Correct Answer is option 2 i.e. 35

Concept

Understanding

Application

Fair calculation

Formula:

x3+y3+z3-3xyz = (x+y+z) (x2+y2+z2-xy-yz-zx)

and

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

Given

x+y+z = 19

xyz = 216

xy+yz+zx = 114

x3+y3+z3-3xyz = (x+y+z) (x2+y2+z2-xy-yz-zx)

⇒ x3+y3+z3-648 = 19((x+y+z)2-2xy-2yz-2xz-xy-yz-zx)

⇒ x3+y3+z3-648 = 19((19)2-3(xy+yz+xz))

⇒ x3+y3+z3-648 = 19(361 – 3 × 114)

⇒ x3+y3+z3-648 = 19 × 19

⇒ x3+y3+z3 = 361 + 648 = 1009

Thus,

 √(x3+y3+z3+xyz)

⇒ √(1009+216)

⇒  √1225 = 35

 

 

 

19(361 – 3 × 114)

= 19(361 – 342)

= 19 × 19