Question

if x+y+z = 6, x²+y²+z²=14 then find the value of x³+y³+z³-3xyz​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: x + y + z = 6 - - - (1)$

and

$\rm \: {x}^{2} + {y}^{2} + {z}^{2} = 14 - - - - (2)$

Now, from equation (1) we have

$\rm \: x + y + z = 6$

On squaring both sides, we get

$\rm \: {(x + y + z)}^{2} = {6}^{2}$

$\rm \: {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx) = 36$

On substituting the value from equation (2), we get

$\rm \: 14 + 2(xy + yz + zx) = 36$

$\rm \: 2(xy + yz + zx) = 36 - 14$

$\rm \: 2(xy + yz + zx) = 22$

$\rm\implies \:xy + yz + zx = 11 - - - - (3)$

Now, Consider

$\rm \: {x}^{3} + {y}^{3} + {z}^{3} - 3xyz$

$\rm \: = (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx)$

$\rm \: = (x + y + z)[{x}^{2} + {y}^{2} + {z}^{2} - (xy + yz + zx) ]$

So, on substituting the values from equation (1), (2) and (3), we get

$\rm \: = \: (6)(14 - 11)$

$\rm \: = \: 6 \times 3$

$\rm \: = \: 18$

Hence,

$\rm\implies \:\boxed{\tt{ \: \: \rm \: {x}^{3} + {y}^{3} + {z}^{3} - 3xyz \: = \: 18 \: \: }}$

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More Identities to know :

➢  (a + b)² = a² + 2ab + b²

➢  (a - b)² = a² - 2ab + b²

➢  a² - b² = (a + b)(a - b)

➢  (a + b)² = (a - b)² + 4ab

➢  (a - b)² = (a + b)² - 4ab

➢  (a + b)² + (a - b)² = 2(a² + b²)

➢  (a + b)³ = a³ + b³ + 3ab(a + b)

➢  (a - b)³ = a³ - b³ - 3ab(a - b)

Answer 2

$\rm\bigstar \large \: Given :$

$\rm\implies \: x + y + z = 6 \: \: (1)$

$\rm\implies \: {x}^{2} + {y}^{2} + {z}^{2} = 14 \: \: (2)$

$\rm \bigstar\large \:To \: Find :$

$\rm\implies \: {x}^{3} + {y}^{3} + {z}^{3} - 3xyz$

$\rm \bigstar\large \: Solution :$

$\rm \: Squaring \: both \: sides \: of \: equation \: (1) \: we \: get$

$\rm (x + y + {z)}^{2} = {(6)}^{2}$

$\rm \implies \: {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx = 36$

$\rm \: By \: using \: '' \: Trinomial \: square \: '' \: identity$

$\rm \implies \: 14+ 2(xy + yz + zx) = 36$

$\rm [ \: by \: using \:(2) ]$

$\rm \implies \: 2(xy + yz + zx) = 36 - 14$

$\rm \implies \: xy + yz + zx= \dfrac{22}{2}$

$\rm \: xy + yz + zx= 11 \: \: (3)$

$\rm \: L. H. S \implies \: {x}^{3} + {y}^{3} + {z}^{3} - 3xyz$

$\rm \: By \: using \: '' \: Special \: Product \: '' \: identity$

$\rm = (x + y + z)[( {x}^{2} + {y}^{2} + {z}^{2} - (xy + yz + zx)]$

$\rm [ \: by \: using \:(1), (2) \: and \: (3)]$

$\rm = (6)[14 - 11] = 6 \times 3 = 18$

$\rm \bigstar \large \: Therefore,$

$\rm \therefore \:The \: required \: solution \: is \: \boxed{ 18}.$