If p2 + q2 - r2 = 0, then the value of (p6 + q6 - r6) ÷ p2q2r2 is:

Question

If p2 + q2 - r2 = 0, then the value of (p6 + q6 - r6) &divide; p2q2r2 is:

Prepp Team · Accepted Answer

Solving the Algebraic Expression Problem

We are given a condition involving $p$, $q$, and $r$, and we need to find the value of a specific algebraic expression.

Given Condition

The given condition is:

\[p^2 + q^2 - r^2 = 0\]

We can rearrange this condition to make it easier to work with:

\[p^2 + q^2 = r^2\]

Expression to Evaluate

We need to find the value of the expression:

\[\frac{p^6 + q^6 - r^6}{p^2q^2r^2}\]

Using Algebraic Identities to Solve

The expression involves terms raised to the power of 6, while the given condition involves terms raised to the power of 2. This suggests we might need to cube the terms in the condition. Let's use the identity for the cube of a sum:

For any two numbers $a$ and $b$, $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$.

From our rearranged condition, we have $p^2 + q^2 = r^2$. Let's cube both sides of this equation:

\[(p^2 + q^2)^3 = (r^2)^3\]

Expanding and Substituting

Using the algebraic identity on the left side, where $a = p^2$ and $b = q^2$:

\[(p^2)^3 + (q^2)^3 + 3(p^2)(q^2)(p^2 + q^2) = r^{2 \times 3}\] \[p^6 + q^6 + 3p^2q^2(p^2 + q^2) = r^6\]

Now, we can substitute the condition $p^2 + q^2 = r^2$ into the equation:

\[p^6 + q^6 + 3p^2q^2(r^2) = r^6\] \[p^6 + q^6 + 3p^2q^2r^2 = r^6\]

Rearranging for the Numerator

We want to find the value of the expression $\frac{p^6 + q^6 - r^6}{p^2q^2r^2}$. Let's rearrange the equation we derived to isolate the numerator $p^6 + q^6 - r^6$:

Subtract $r^6$ from both sides:

\[p^6 + q^6 + 3p^2q^2r^2 - r^6 = r^6 - r^6\] \[p^6 + q^6 - r^6 + 3p^2q^2r^2 = 0\]

Subtract $3p^2q^2r^2$ from both sides:

\[p^6 + q^6 - r^6 = -3p^2q^2r^2\]

Evaluating the Expression

Now we can substitute this result back into the expression we need to evaluate:

\[\frac{p^6 + q^6 - r^6}{p^2q^2r^2} = \frac{-3p^2q^2r^2}{p^2q^2r^2}\]

Assuming $p^2q^2r^2 \neq 0$, we can cancel the $p^2q^2r^2$ terms from the numerator and the denominator:

\[\frac{-3\cancel{p^2q^2r^2}}{\cancel{p^2q^2r^2}} = -3\]

So, the value of the expression is -3.

Let's quickly look at the options provided:

-3
1/3
3
3pqr

Our calculated value is -3, which matches the first option.

Revision Table: Key Algebraic Concepts

Concept	Description	Relevance to Problem
Algebraic Expression	A mathematical phrase that contains numbers, variables, and operators.	The problem involves evaluating an expression.
Algebraic Identity	An equation that is true for all possible values of the variables involved (e.g., $(a+b)^3$ identity).	Used to simplify the given condition and relate powers of 2 to powers of 6.
Substitution	Replacing a variable or expression with its equivalent value.	Used repeatedly to simplify equations and find the final value.

Additional Information: Algebraic Identities

Algebraic identities are powerful tools for simplifying expressions and solving equations. Here are a few common ones:

Square of a sum: $(a+b)^2 = a^2 + 2ab + b^2$
Square of a difference: $(a-b)^2 = a^2 - 2ab + b^2$
Difference of squares: $a^2 - b^2 = (a-b)(a+b)$
Cube of a sum: $(a+b)^3 = a^3 + b^3 + 3ab(a+b) = a^3 + b^3 + 3a^2b + 3ab^2$
Cube of a difference: $(a-b)^3 = a^3 - b^3 - 3ab(a-b) = a^3 - b^3 - 3a^2b + 3ab^2$
Sum of cubes: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
Difference of cubes: $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

Another important identity used when $a+b+c=0$ is $a^3+b^3+c^3=3abc$. While we didn't use this directly here due to the structure of the problem ($p^2+q^2 = r^2$, not $p^2+q^2+(-r^2)=0$ when considering the cubes), the approach of cubing the sum was effective.

Prepp Team · Answer

1/3

Prepp Team · Answer

3

Prepp Team · Answer

3pqr