Question

7th question which is in the attachment...pls answer it fast..

Answer 1

Given that,

$\displaystyle\longrightarrow\sf{\left|\begin {array}{ccc}\sf{x}&\sf{x^2}&\sf{1+x^3}\\\sf{y}&\sf{y^2}&\sf{1+y^3}\\\sf{z}&\sf{z^2}&\sf{1+z^3}\end {array}\right|=0}$

Taking the determinant,

$\displaystyle\longrightarrow\sf{x[y^2(1+z^3)-z^2(1+y^3)]-x^2[y(1+z^3)-z(1+y^3)]+(1+x^3)[yz^2-y^2z]=0}$

We have to expand each term.

$\displaystyle\longrightarrow\sf{x(y^2+y^2z^3-z^2-y^3z^2)-x^2(y+yz^3-z-y^3z)+yz^2-y^2z+x^3yz^2-x^3y^2z=0}$

$\displaystyle\longrightarrow\sf{xy^2+xy^2z^3-xz^2-xy^3z^2-x^2y-x^2yz^3+x^2z+x^2y^3z+yz^2-y^2z+x^3yz^2-x^3y^2z=0}$

$\displaystyle\longrightarrow\sf{xy^2z^3-xy^3z^2-x^2yz^3+x^2y^3z+x^3yz^2-x^3y^2z=x^2y-xy^2+xz^2-x^2z+y^2z-yz^2}$

Take $\displaystyle\sf {xyz}$ common in LHS.

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

$\displaystyle\longrightarrow\sf{xyz=\dfrac {x^2y-xy^2+xz^2-x^2z+y^2z-yz^2}{yz^2-y^2z+x^2z-xz^2+xy^2-x^2y}}$

Finally,

$\displaystyle\longrightarrow\sf{\underline {\underline {xyz=-1}}}$

Done!