Q1(viii):

Check whether the following are quadratic equations : (viii) $x^3 – 4x^2 – x + 1 = (x – 2)^3$

Given: The equation $(x^3 - 4x^2 - x + 1) = (x - 2)^3$.

To Find: Determine whether the given equation is a quadratic equation.

Step 1: Understanding the definition of a Quadratic Equation
A quadratic equation is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$. The highest power (degree) of the variable $x$ must be exactly 2.

Step 2: Expanding the right-hand side (RHS)
The RHS is $(x - 2)^3$. We use the algebraic identity for the cube of a binomial:
$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
Here, $a = x$ and $b = 2$.
$(x - 2)^3 = x^3 - 3(x^2)(2) + 3(x)(2^2) - (2)^3$
$(x - 2)^3 = x^3 - 6x^2 + 3(x)(4) - 8$
$(x - 2)^3 = x^3 - 6x^2 + 12x - 8$ [Applying the identity $(a-b)^3$]

Step 3: Equating LHS and RHS
Substitute the expanded RHS back into the original equation:
$x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$

Step 4: Simplifying the equation
To simplify, we bring all terms to the left-hand side (LHS):
$(x^3 - 4x^2 - x + 1) - (x^3 - 6x^2 + 12x - 8) = 0$
$x^3 - 4x^2 - x + 1 - x^3 + 6x^2 - 12x + 8 = 0$ [Distributing the negative sign]

Step 5: Combining like terms
Group the terms by their powers of $x$:
$(x^3 - x^3) + (-4x^2 + 6x^2) + (-x - 12x) + (1 + 8) = 0$
$0x^3 + 2x^2 - 13x + 9 = 0$
$2x^2 - 13x + 9 = 0$

Step 6: Conclusion
The resulting equation is $2x^2 - 13x + 9 = 0$. This equation is in the form $ax^2 + bx + c = 0$, where $a = 2$, $b = -13$, and $c = 9$. Since the highest degree of the variable $x$ is 2 and $a \neq 0$, the equation satisfies the definition of a quadratic equation.

Final Answer: Yes, the given equation is a quadratic equation.