Math, asked by killer6164, 8 months ago

if x+y+z =6 x^2 + y^2 +z^2 = 18 then find x^3 + y^3 + z^3​

Answers

Answered by anindyaadhikari13
2

\star\:\:\:\sf\large\underline\blue{Question:-}

  • If \sf x+y+z=6and \sf x^{2}+y^{2}+z^{2}=18, find the value of \sf x^{3}+y^{3}+z^{3}

\star\:\:\:\sf\large\underline\blue{Solution:-}

Given,

\sf x + y + z = 6

  \sf{x}^{2} +  {y}^{2}  +  {z}^{2}  = 18

Now,

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

 \sf \implies {x}^{2}  +  {y}^{2}  +  {z}^{2}  + 2(xy + yz + xz) = 36

 \sf \implies 18  + 2(xy + yz + xz) = 36

 \sf \implies  2(xy + yz + xz) = 18

 \sf \implies  (xy + yz + xz) = 9

Now,

 \sf {x}^{3}  +  {y}^{3}  +  {z}^{3}

 \sf = (x + y + z)( {x}^{2}  +  {y}^{2}  +  {z}^{2}  - xy - yz - xz)

 \sf = 6 \times (18 - 9)

 \sf = 6  \times 9

 \sf = 54

\star\:\:\:\sf\large\underline\blue{Answer:-}

  •  \sf {x}^{3}  +  {y}^{3}  +  {z}^{3}  = 54
Answered by nehashanbhag0729
0

Answer:

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