Question
If (1/x) + (1/y) + (1/z) = 0 and x + y + z = 9, then what is the value of x^3 + y^3 + z^3 – 3xyz?
यदि (1 / x) + (1 / y) + (1 / z) = 0 और x + y + z = 9, तो x^3 + y^3 + z^3 - 3xyz का मान क्या है?
Answer C.
C.1/x + 1/y +1/z = 0
yz + xz + xy/ xyz = 0
yz + xz + xy = 0-----(I)
Given-
x + y + z = 9-----(II)
Squaring both sides
(x + y + z)² = (9)²
x² + y² + z² + 2xy + 2yz + 2xz = 81
x² + y² + z² + 2(xy + yz + xz) = 81
x² + y² + z² + 2(0) = 81 [from equation 1.(x+y+z=0)]
x² + y² + z² = 81 ----- (III)
Now- x³ + y³ + z³ - 3xyz = (x+y+z){x² + y² + z² - (xy + yz + zx)}
Now put the values from equation (1) (2) and (3)
x³ + y³ + z³ - 3xyz = (9)(81 - 0)
x³ + y³ + z³ - 3xyz = 729
So the correct answer is option C.
C.1 / x + 1 / y + 1 / z = 0
yz + xz + xy / xyz = 0
yz + xz + xy = 0 ----- (I)
दिया हुआ है -
x + y + z = 9 ----- (II)
दोनों पक्षों में वर्ग करने पर
(x + y + z)² = (9)²
x² + y² + z² + 2xy + 2yz + 2xz = 81
x² + y² + z² + 2(xy + yz + xz) = 81
x² + y² + z² + 2(0) = 81 [समीकरण 1 से x + y + z = 9 ]
x² + y² + z² = 81 ----- (III)
अब-
x³ + y³ + z³ - 3xyz = (x+y+z){x² + y² + z² - (xy + yz + zx)}
अब समीकरण (1) (2) और (3) से मान रखने पर-
x³ + y³ + z³ - 3xyz = (9)(81 - 0)
x³ + y³ + z³ - 3xyz = 729
इसलिए सही उत्तर विकल्प C है।
