If \(\mathrm{a}^{2}-4 \mathrm{a}-1=0, \mathrm{a} \neq 0\), then the value of \(\mathrm{a}^{2}+3 \mathrm{a}+\frac{1}{\mathrm{a}^{2}}-\frac{3}{\mathrm{a}}\) is (a) 24 (b) 26 (c) 28 (d) 30

Given the system of equations: 1. \(a + b + c = 3\) 2. \(a^2 + b^2 + c^2 = 6\) 3. \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\) We want to find the value of \(abc\). To solve this, let's first analyze the given information. The third equation can be written as: \[ \frac{ab + ac + bc}{abc} = 1 \] FRead more

Given the system of equations:

1. \(a + b + c = 3\)

2. \(a^2 + b^2 + c^2 = 6\)

3. \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\)

We want to find the value of \(abc\).

To solve this, let’s first analyze the given information. The third equation can be written as:

\[

\frac{ab + ac + bc}{abc} = 1

\]

From this, we get:

\[

ab + ac + bc = abc

\]

From the first equation, we know the sum of \(a\), \(b\), and \(c\). To find \(ab + ac + bc\), we can square the first equation and compare it to the second equation. Squaring the first equation gives us:

\[

(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) = 3^2 = 9

\]

Given \(a^2 + b^2 + c^2 = 6\) from the second equation, we can substitute that into our squared first equation to find \(ab + ac + bc\):

\[

6 + 2(ab + ac + bc) = 9

\]

Solving for \(ab + ac + bc\):

\[

2(ab + ac + bc) = 3

\]

\[

ab + ac + bc = \frac{3}{2}

\]

Given \(ab + ac + bc = abc\), we have:

\[

abc = \frac{3}{2}

\]

Therefore, the value of \(abc\) is \(\frac{3}{2}\).

**Answer: (b) \(\frac{3}{2}\)**.

Given \[ \begin{aligned} & a^2-4 a-1=0 \\ & a^2-4 a=1 \\ & a(a-4)=1 \\ & a-4=\frac{1}{a} \\ & a-\frac{1}{a}=4 \end{aligned} \] We have \(\mathrm{a}^2+3 \mathrm{a}+\frac{1}{\mathrm{a}^2}-\frac{3}{\mathrm{a}}\) \[ \begin{aligned} & \left(a^2+\frac{1}{a^2}\right)+3\left(a-\frac{Read more

Given

\[

\begin{aligned}

& a^2-4 a-1=0 \\

& a^2-4 a=1 \\

& a(a-4)=1 \\

& a-4=\frac{1}{a} \\

& a-\frac{1}{a}=4

\end{aligned}

\]

We have \(\mathrm{a}^2+3 \mathrm{a}+\frac{1}{\mathrm{a}^2}-\frac{3}{\mathrm{a}}\)

\[

\begin{aligned}

& \left(a^2+\frac{1}{a^2}\right)+3\left(a-\frac{1}{a}\right) \\

& \left(a-\frac{1}{a}\right)^2+3\left(a-\frac{1}{a}\right)+2 \\

& 4^2+3 \times 4+2=30

\end{aligned}

\]

Therefore, Correct option is (d).

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