Let's mix it up

Algebra Level 5

$\triangle =\begin{vmatrix} x & { x }^{ 2 } & 1+{ x }^{ 3 } \\ y & y^{ 2 } & { 1+y }^{ 3 } \\ z & z^{ 2 } & 1+z^{ 3 } \end{vmatrix}=0$ where $x\neq y\neq z$ .

Consider a determinant above.

Now find the value of $x\times y\times z$ and let it be equal to $A$ , where $A$ is an integer.

Now consider a binomial $({ \sqrt [ 4 ]{ 6 } +\sqrt [ 6 ]{ 4 } ) }^{ 50 }$

Let the number of rational terms in the expansion of above binomial is $B$ and the number of irrational terms be $C$ .

Now if, it is given that $\frac { C-\lambda B - 1}{ 2 } =-A$ where $\lambda$ is a constant. Then find the value of $B+C+\lambda - 1 - A$ .

Hint: Use properties of determinant to solve the determinant.

The answer is 62.

1 solution

Yash Choudhary
Apr 2, 2015

$\Rightarrow \quad \begin{vmatrix} x & { x }^{ 2 } & 1+{ x }^{ 3 } \\ y & { y }^{ 2 } & 1+{ y }^{ 3 } \\ z & { z }^{ 2 } & 1+{ z }^{ 3 } \end{vmatrix}=0\\ \Rightarrow \quad \begin{vmatrix} x & { x }^{ 2 } & 1 \\ y & { y }^{ 2 } & 1 \\ z & { z }^{ 2 } & 1 \end{vmatrix}+\begin{vmatrix} x & { x }^{ 2 } & { x }^{ 3 } \\ y & { y }^{ 2 } & { y }^{ 3 } \\ z & { z }^{ 2 } & { z }^{ 3 } \end{vmatrix}=0\\ \Rightarrow \quad \begin{vmatrix} x & { x }^{ 2 } & 1 \\ y & { y }^{ 2 } & 1 \\ z & { z }^{ 2 } & 1 \end{vmatrix}+(xyz)\begin{vmatrix} x & { x }^{ 2 } & 1 \\ y & { y }^{ 2 } & 1 \\ z & { z }^{ 2 } & 1 \end{vmatrix}=0\\ \Rightarrow \quad \begin{vmatrix} x & { x }^{ 2 } & 1 \\ y & { y }^{ 2 } & 1 \\ z & { z }^{ 2 } & 1 \end{vmatrix}+(xyz)\begin{vmatrix} x & { x }^{ 2 } & 1 \\ y & { y }^{ 2 } & 1 \\ z & { z }^{ 2 } & 1 \end{vmatrix}=0\\ \Rightarrow \quad (\begin{vmatrix} x & { x }^{ 2 } & 1 \\ y & { y }^{ 2 } & 1 \\ z & { z }^{ 2 } & 1 \end{vmatrix})(xyz+1)=0\\ \therefore \quad xyz=-1=A\\ \Rightarrow \quad { (\sqrt [ 4 ]{ 6 } +\sqrt [ 6 ]{ 4 } ) }^{ 50 }\\ \Rightarrow \quad { T }_{ r+1 }={ ^{ 50 }C }_{ r }{ (6) }^{ ^{ \frac { 50-r }{ 4 } } }{ (4) }^{ \frac { r }{ 6 } }\\ \therefore \quad Rational\quad terms\quad are\quad for\quad r=6,18,30,42\\ \Rightarrow \quad B=4\quad and\quad C=47\\ \Rightarrow \quad Solving\quad further\quad equation\quad we\quad get\quad \lambda =11\\ \Rightarrow \quad B+C+\lambda-A-1 =\boxed { 62 } .$

Is it just me or is this sum way too overrated???

Rushikesh Joshi - 6 years, 2 months ago

too easy for level 5...

Vivek Modi - 6 years, 2 months ago

The value of $\lambda$ comes out to be 11 not 12 :)It may be a typo. Anyways nice solution… +1

Aditya Tiwari - 6 years, 2 months ago

Ohhh yeah. That was a mistake, I corrected it.

Yash Choudhary - 6 years, 2 months ago

Let's mix it up

The answer is 62.

1 solution

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