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Question Number 11272 by chux last updated on 18/Mar/17

Answered by mrW1 last updated on 18/Mar/17

(1)⇒y+z=1−x (2)⇒y^2 +z^2 =35−x^2 (y+z)^2 =(1−x)^2 y^2 +z^2 +2yz=1+x^2 −2x 35−x^2 +2yz=1+x^2 −2x yz=x^2 −x−17 (3)⇒y^3 +z^3 =97−x^3 (y+z)(y^2 +z^2 −yz)=97−x^3 (1−x)(35−x^2 −x^2 +x+17)=97−x^3 (1−x)(52−2x^2 +x)=97−x^3 52−2x^2 +x−52x+2x^3 −x^2 −97+x^3 =0 −3x^2 −51x+3x^3 −45=0 x^3 −x^2 −17x−15=0 (x+1)(x+3)(x−5)=0 ⇒x=−3, −1, 5 similarly ⇒y=−3, −1, 5 ⇒z=−3, −1, 5 all possible solutions are therefore ((x),(y),(z) )= (((−3)),((−1)),(5) ) ((x),(y),(z) )= (((−3)),(5),((−1)) ) ((x),(y),(z) )= (((−1)),((−3)),(5) ) ((x),(y),(z) )= (((−1)),(5),((−3)) ) ((x),(y),(z) )= ((5),((−3)),((−1)) ) ((x),(y),(z) )= ((5),((−1)),((−3)) )

(1)⇒y+z=1−x(2)⇒y2+z2=35−x2(y+z)2=(1−x)2y2+z2+2yz=1+x2−2x35−x2+2yz=1+x2−2xyz=x2−x−17(3)⇒y3+z3=97−x3(y+z)(y2+z2−yz)=97−x3(1−x)(35−x2−x2+x+17)=97−x3(1−x)(52−2x2+x)=97−x352−2x2+x−52x+2x3−x2−97+x3=0−3x2−51x+3x3−45=0x3−x2−17x−15=0(x+1)(x+3)(x−5)=0⇒x=−3,−1,5similarly⇒y=−3,−1,5⇒z=−3,−1,5allpossiblesolutionsaretherefore(xyz)=(−3−15)(xyz)=(−35−1)(xyz)=(−1−35)(xyz)=(−15−3)(xyz)=(5−3−1)(xyz)=(5−1−3)

Commented by chux last updated on 19/Mar/17

i really appreciate ....thanx alot.

ireallyappreciate....thanxalot.

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