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Grade 11Discuss with colleagues and IITians

If x+y+z=1 Then (1-3x^2-3y^2-3z^2+2x^3+2y^3+2z^3)=? Please provide the solution

Shrey

12 Years agoGrade 11

2 Answers

Parvez ali

12 Years ago

Thanks & Regards

Parvez Ali

askIITians Faculty

Sher Mohammad

12 Years ago

$x+y+z=1$

$1-3x^2-3y^2-3z^2+2x^3+2y^3+2z^3$

(x+y+z)^2=(

(x+y+z)^3=

2(x+y+z)^2-3(x+y+z)^2=-1=

$1-3x^2-3y^2-3z^2+2x^3+2y^3+2z^3$

from above equations,

$1-3x^2-3y^2-3z^2+2x^3+2y^3+2z^3=6xyz$

Sher Mohammad

B.Tech, IIT Delhi

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