If [tex]a = x^{q+r} \cdot y^p[/tex], [tex]b = x^{r+p} \cdot y^q[/tex], [tex]c = x^{p+q} \cdot y^r[/tex], prove that [tex]a^{q-r} \times b^{r-p} \times c^{p-q} = 1[/tex].



Answer :

To explore and simplify the given expression [tex]\(a^{q-r} \times b^{r-p} \times c^{p-q}\)[/tex], let’s consider each element step by step.

First, recall the definitions:

[tex]\[ a = x^{q+r} \cdot y^p, \quad b = x^{r+p} \cdot y^q, \quad c = x^{p+q} \cdot y^r \][/tex]

The overall goal is to simplify the product [tex]\(a^{q-r} \times b^{r-p} \times c^{p-q}\)[/tex].

### Step 1: Identify Individual Powers

For [tex]\(a\)[/tex]:

[tex]\[ a^{q-r} = (x^{q+r} \cdot y^p)^{q-r} \][/tex]

Applying the power to each term inside the parentheses:

[tex]\[ a^{q-r} = x^{(q+r)(q-r)} \cdot y^{p(q-r)} \][/tex]

For [tex]\(b\)[/tex]:

[tex]\[ b^{r-p} = (x^{r+p} \cdot y^q)^{r-p} \][/tex]

Again, applying the power to each term:

[tex]\[ b^{r-p} = x^{(r+p)(r-p)} \cdot y^{q(r-p)} \][/tex]

For [tex]\(c\)[/tex]:

[tex]\[ c^{p-q} = (x^{p+q} \cdot y^r)^{p-q} \][/tex]

Applying the power to each term:

[tex]\[ c^{p-q} = x^{(p+q)(p-q)} \cdot y^{r(p-q)} \][/tex]

### Step 2: Multiply the Simplified Terms

Combine the three terms:

[tex]\[ a^{q-r} \times b^{r-p} \times c^{p-q} = (x^{(q+r)(q-r)} \cdot y^{p(q-r)}) \times (x^{(r+p)(r-p)} \cdot y^{q(r-p)}) \times (x^{(p+q)(p-q)} \cdot y^{r(p-q)}) \][/tex]

### Step 3: Simplify Using Properties of Exponents

Combine the exponents for the [tex]\(x\)[/tex] terms:

[tex]\[ x^{(q+r)(q-r) + (r+p)(r-p) + (p+q)(p-q)} \][/tex]

Combine the exponents for the [tex]\(y\)[/tex] terms:

[tex]\[ y^{p(q-r) + q(r-p) + r(p-q)} \][/tex]

### Step 4: Evaluate the Exponent Expressions

1. Simplify the [tex]\(x\)[/tex] term's exponent:

[tex]\[ (q+r)(q-r) + (r+p)(r-p) + (p+q)(p-q) \][/tex]

Using the difference of squares:

[tex]\[ = (q^2 - r^2) + (r^2 - p^2) + (p^2 - q^2) \][/tex]

These terms cancel each other out:

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

Thus:

[tex]\[ x^0 = 1 \][/tex]

2. Simplify the [tex]\(y\)[/tex] term's exponent:

[tex]\[ p(q-r) + q(r-p) + r(p-q) \][/tex]

Distribute and combine like terms:

[tex]\[ = pq - pr + qr - pq + rp - rq \][/tex]

Here again, all terms cancel each other out:

[tex]\[ = pq - pq + qr - rq + rp - pr = 0 \][/tex]

Thus:

[tex]\[ y^0 = 1 \][/tex]

### Conclusion

Combining the results, we have:

[tex]\[ a^{q-r} \times b^{r-p} \times c^{p-q} = 1 \cdot 1 = 1 \][/tex]

However, when verified correctly, we find that all assumptions were valid, and exponents should cancel correctly. Given the result is true yet seems counter-intuitive, it concludes the expression should ideally simplify to [tex]\(1\)[/tex], indicating the result is logically derived despite practical verification sometimes giving nuances.