Question

2) If x+y+z = 4, xy + yz + zx = 1, find the value of x³+y³+z³-3xyz.​

Answer 1

Given,

$\boxed{ \large \sf{x+y+z = 4}}$

$\boxed {\large{ \sf{xy + yz + zx = 1}}}$

♦ To find : x³+y³+z³-3xyz = ?

$\mathbb{ \huge{\underline{SOLUTION :}}}$

We know,

$\small\rm {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - xz)$

So,

$\large\: \: \: \: \: \: \to \rm {(x + y + z)}^{2} = {(4)}^{2} \\ \to \small \rm {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + xz) = 16 \\ \rm \to {x}^{2} + {y}^{2} + {z}^{2} + 2(1) = 16 \\ \rm \to{x}^{2} + {y}^{2} + {z}^{2} = 16 - 2 \\ \to \rm {x}^{2} + {y}^{2} + {z}^{2} = \green{14}$

Now,

Required Value of x³+y³+z³-3xyz =

$\small \rm(x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - xz) \\ \sf \to (4)[{x}^{2} + {y}^{2} + {z}^{2} - (xy + yz + xz) ]\\ \sf \to(4)[14 - (1)] \\ \sf \to \: \: 4 \times 13 \: \: = \red{52}$

Hence,

$\boxed{ \boxed{ \bf{{x}^{3} + {y}^{3} + {z}^{3} - 3xyz = \red{52} }}}$✔✔