If x + y + z = 1, and xy + yz + zx = -1 and xyz = -1, then find the value of $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ .
Answer
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Hint: In these types of questions algebraic identities play an important role for the solution part. Here we will see and use some important algebraic identities and try to solve the algebraic expressions using another algebraic expression. Also, the Product of two negative numbers is positive.
Complete step-by-step answer:
We are given that x + y + z = 1 ….. ( i ) , xy +yz +zx =-1……. ( ii ) , xyz = -1 ….. ( iii )
Now, we know that \[{{(~x\text{ }+~y\text{ }+z\text{ })}^{2}}\text{ }={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2(xy+yz+zx)\] …… ( iv )
Putting ( i ) and ( ii ) in ( iv ), we get
\[{{(~-1)}^{2}}\text{ }={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2(-1)\]
\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=3\]…… ( v )
We know another identity which is $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ - 3xyz = ( x + y + z )( \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx\])
Re-writing above equation, $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ - 3(xyz) = ( x + y + z )( \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-(xy+yz+zx)\]) ….. ( vi )
Now, we can see that we have all the values of equations except $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ ,
So, putting equation ( i ), in equation ( vi ), we get
$ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ - 3(xyz) = ( 1 )( \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-(xy+yz+zx)\])
, putting equation ( ii ), in equation ( vi ), we get
$ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ - 3(xyz) = ( 1 )( \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-(-1)\])
putting equation ( iii ) in equation ( vi ), we get
$ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ - 3 ( - 1 ) = ( 1 ) ( \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-(-1)\])
putting equation ( v ) in equation ( vi ), we get
$ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ - 3 ( - 1 ) = ( 1 ) ( \[3-(-1)\])
On simplifying, we get
$ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ + 3 = 4
On solving numerical values we get,
$ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ = 1
Hence, the value of $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ is equals to 1 .
Note: We have to use only those algebraic identities which will help us in solving part and reducing algebraic form into the simplest form. Calculation should be done very carefully. Expansion of algebraic expression must be correct and the sign scheme must be properly checked once as this may lead to wrong answers.
Complete step-by-step answer:
We are given that x + y + z = 1 ….. ( i ) , xy +yz +zx =-1……. ( ii ) , xyz = -1 ….. ( iii )
Now, we know that \[{{(~x\text{ }+~y\text{ }+z\text{ })}^{2}}\text{ }={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2(xy+yz+zx)\] …… ( iv )
Putting ( i ) and ( ii ) in ( iv ), we get
\[{{(~-1)}^{2}}\text{ }={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2(-1)\]
\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=3\]…… ( v )
We know another identity which is $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ - 3xyz = ( x + y + z )( \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx\])
Re-writing above equation, $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ - 3(xyz) = ( x + y + z )( \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-(xy+yz+zx)\]) ….. ( vi )
Now, we can see that we have all the values of equations except $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ ,
So, putting equation ( i ), in equation ( vi ), we get
$ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ - 3(xyz) = ( 1 )( \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-(xy+yz+zx)\])
, putting equation ( ii ), in equation ( vi ), we get
$ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ - 3(xyz) = ( 1 )( \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-(-1)\])
putting equation ( iii ) in equation ( vi ), we get
$ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ - 3 ( - 1 ) = ( 1 ) ( \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-(-1)\])
putting equation ( v ) in equation ( vi ), we get
$ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ - 3 ( - 1 ) = ( 1 ) ( \[3-(-1)\])
On simplifying, we get
$ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ + 3 = 4
On solving numerical values we get,
$ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ = 1
Hence, the value of $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ is equals to 1 .
Note: We have to use only those algebraic identities which will help us in solving part and reducing algebraic form into the simplest form. Calculation should be done very carefully. Expansion of algebraic expression must be correct and the sign scheme must be properly checked once as this may lead to wrong answers.
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