The HCF of $x$ and $y$ is $H$. Consider the following statements in respect of the HCF of $p=\frac{x^3 + y^3}{x^2-xy+y^2}$ and $q=\frac{x^3-y^3}{x^2+xy+y^2}$ : I. The HCF of $p$ and $q$ can be $H$. II. The HCF of $p$ and $q$ can be $2H$. Which of the statements given above is/are correct?

Question

Prepp Team · Accepted Answer

HCF of Algebraic Expressions Analysis

The question asks us to determine which statements are correct regarding the Highest Common Factor (HCF) of two algebraic expressions, $p$ and $q$, given that the HCF of $x$ and $y$ is $H$.

First, let's simplify the expressions for $p$ and $q$. We are given:

$p = \frac{x^3 + y^3}{x^2 - xy + y^2}$
$q = \frac{x^3 - y^3}{x^2 + xy + y^2}$

We know the algebraic identities for the sum and difference of cubes:

$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$
$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

Simplifying Expression p

Substitute the sum of cubes identity into the expression for $p$:

$p = \frac{(x + y)(x^2 - xy + y^2)}{x^2 - xy + y^2}$

Assuming $x^2 - xy + y^2 \neq 0$, we can cancel the term $x^2 - xy + y^2$ from the numerator and the denominator:

$p = x + y$

Simplifying Expression q

Substitute the difference of cubes identity into the expression for $q$:

$q = \frac{(x - y)(x^2 + xy + y^2)}{x^2 + xy + y^2}$

Assuming $x^2 + xy + y^2 \neq 0$, we can cancel the term $x^2 + xy + y^2$ from the numerator and the denominator:

$q = x - y$

Finding HCF of Simplified Expressions

Now we need to find the HCF of $p = x + y$ and $q = x - y$. We are given that HCF($x, y$) = $H$.

This means we can write $x$ and $y$ as:

$x = H \cdot a$

$y = H \cdot b$

where $a$ and $b$ are coprime integers, meaning HCF($a, b$) = 1.

Substitute these into the simplified expressions for $p$ and $q$:

$p = Ha + Hb = H(a + b)$

$q = Ha - Hb = H(a - b)$

The HCF of $p$ and $q$ is HCF($H(a+b), H(a-b)$). We can factor out $H$:

HCF($p, q$) = $H \cdot$ HCF($a+b, a-b$)

Analyzing HCF(a+b, a-b)

Let $d$ be the HCF of $a+b$ and $a-b$. So, $d =$ HCF($a+b, a-b$).

Since $d$ divides both $a+b$ and $a-b$, it must also divide their sum and difference:

$d$ divides $(a+b) + (a-b) = 2a$.
$d$ divides $(a+b) - (a-b) = 2b$.

Therefore, $d$ is a common divisor of $2a$ and $2b$. This implies that $d$ must divide the HCF of $2a$ and $2b$.

HCF($2a, 2b$) = $2 \cdot$ HCF($a, b$)

Since we know HCF($a, b$) = 1, we have:

HCF($2a, 2b$) = $2 \cdot 1 = 2$

So, $d$ must divide 2. The possible integer values for $d$ are 1 and 2.

Evaluating Statement I: HCF can be H

This statement is correct if HCF($a+b, a-b$) can be 1.

Let's test if $d=1$ is possible. We need $a$ and $b$ such that HCF($a, b$) = 1 and HCF($a+b, a-b$) = 1.

Consider $a=2$ and $b=1$. HCF($a, b$) = HCF(2, 1) = 1.

Then $a+b = 2+1 = 3$ and $a-b = 2-1 = 1$.

HCF($a+b, a-b$) = HCF(3, 1) = 1.

In this case, HCF($p, q$) = $H \cdot$ HCF($a+b, a-b$) = $H \cdot 1 = H$.

So, the HCF of $p$ and $q$ can indeed be $H$. Statement I is correct.

Evaluating Statement II: HCF can be 2H

This statement is correct if HCF($a+b, a-b$) can be 2.

For $d=2$, both $a+b$ and $a-b$ must be even. This happens when $a$ and $b$ have the same parity (both even or both odd).

Since HCF($a, b$) = 1, $a$ and $b$ cannot both be even. Therefore, they must both be odd.

Let's test if $d=2$ is possible with $a$ and $b$ being odd. Consider $a=3$ and $b=1$. HCF($a, b$) = HCF(3, 1) = 1.

Then $a+b = 3+1 = 4$ and $a-b = 3-1 = 2$.

HCF($a+b, a-b$) = HCF(4, 2) = 2.

In this case, HCF($p, q$) = $H \cdot$ HCF($a+b, a-b$) = $H \cdot 2 = 2H$.

So, the HCF of $p$ and $q$ can indeed be $2H$. Statement II is correct.

Conclusion on Statements

Since both statement I (HCF can be $H$) and statement II (HCF can be $2H$) are possible under the given conditions, both statements are correct.

Prepp Team · Answer

I only

Prepp Team · Answer

II only

Prepp Team · Answer

Neither I nor II

HCF of Algebraic Expressions Analysis

Simplifying Expression p

Simplifying Expression q

Finding HCF of Simplified Expressions

Analyzing HCF(a+b, a-b)

Evaluating Statement I: HCF can be H

Evaluating Statement II: HCF can be 2H

Conclusion on Statements

Similar Questions

Important Questions from LCM and HCF

UPSC CDS 2026 Mock Test Series

CDS Mock Test Series

Start Your Preparation with Prepp Mobile App