Q1(vii):

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

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

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

Definition: A quadratic equation in the variable $x$ 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 1: Expanding the Left-Hand Side (LHS)
The LHS 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 $b = 2$.
$(x + 2)^3 = x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3$
$= x^3 + 6x^2 + 3(x)(4) + 8$
$= x^3 + 6x^2 + 12x + 8$ [Expanding the terms]

Step 2: Expanding the Right-Hand Side (RHS)
The RHS is $2x(x^2 - 1)$. We use the distributive property of multiplication over subtraction: $a(b - c) = ab - ac$.
$2x(x^2 - 1) = (2x \cdot x^2) - (2x \cdot 1)$
$= 2x^3 - 2x$ [Distributing $2x$ into the parentheses]

Step 3: Equating LHS and RHS and Simplifying
Now, set the expanded LHS equal to the expanded RHS:
$x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$
To bring all terms to one side, subtract $(2x^3 - 2x)$ from both sides:
$x^3 - 2x^3 + 6x^2 + 12x + 2x + 8 = 0$
$-x^3 + 6x^2 + 14x + 8 = 0$ [Combining like terms]

Step 4: Analyzing the Degree of the Equation
The resulting equation is $-x^3 + 6x^2 + 14x + 8 = 0$.
The highest power (degree) of the variable $x$ in this equation is 3. [Since the term $-x^3$ exists and its coefficient is non-zero].

Conclusion:
Since the degree of the equation is 3, it is a cubic equation, not a quadratic equation. A quadratic equation must have a maximum degree of 2.

Final Answer: The given equation $(x + 2)^3 = 2x(x^2 - 1)$ is not a quadratic equation.