दिया हैं,
`|(x,x^(2),1+x^(3)),(2x,4x^(2),1+ 8x^(3)),(3x,9x^(2),1+ 27x^(3))| = 10`
अतः ` C_1` से `x` और `C_2` से `x^2` उभयनिष्ट लेने पर,
`x xx x^(2) |(1,1,1+x^(3)),(2,4,1+ 8x^(3)),(3,9,1+ 27x^(3))| = 10`
संक्रिया `R_2rightarrowR_2-2R_1` से,
`x^(3) |(1,1,1+x^(3)),(0,2,-1+6x^(3)),(3,9,1+ 27x^(3))| = 10`
संक्रिया `R_3rightarrowR_3-3R_1` से,
`x^(3) |(1,1,1+x^(3)),(0,2,-1+6x^(3)),(0,6,-2 + 24x^(3))| = 10`
`C_1` से प्रसार करने पर,
`x^(3)[1{2(-2 + 24x^(3)) - 6( -1+6x^(3))}]= 10`
`implies x^(3)[-4 + 48x^(3)- (-6+36^(3 )) ]= 10`
`implies x^(3)[-4 + 48x^(3) + 6-36^(3 ) ]= 10`
`implies x^(3)[2 + 12x^(3 ) ]= 10`
`implies 2x^(3) + 12x^(6 ) = 10`
2 उभयनिष्ठ लेने पर,
`implies 2(x^(3) + 6x^(6 )) = 10 `
`implies x^(3) + 6x^(6 ) = 10/2 `
`implies x^(3) + 6x^(6 ) = 5`
माना की `x^(3) = y`
अतः `y + 6y^(2 ) = 5`
`implies6y^(2 )+y - 5 = 0`
`implies6y^(2 )+(6y-5y ) - 5 = 0`
`implies6y^(2 )+(6y-5y ) - 5 = 0`
`implies6y^(2 )+6y-5y - 5 = 0`
`implies6y (y+1 )+ (-5) (y+1 )= 0`
`(y+1 )` उभयनिष्ठ लेने पर,
`implies(6y-5 )+ (y+1 )= 0`
अब, `6y - 5 = 0` और `y+1 = 0`
`implies6y = 5` और `y = -1`
चूँकि `x^(3) = y`
`6x ^(3 ) = 5`, और `x ^(3 ) = -1`
`impliesx ^(3 ) = 5/6`, और `x ^(3 ) = -1`
`impliesx = (5/6) ^(1/3 )`, और `x ^(3 ) = -1`
अतः `x = root(3)(5/6)`और `x = -1 `
अतः `x` के लिए हलों की संख्या `2` हैं|
