Certainly! Let's prove the given expression step-by-step.
### Given:
- [tex]\( a = x^{q+r} \cdot y^p \)[/tex]
- [tex]\( b = x^{r+p} \cdot y^q \)[/tex]
- [tex]\( c = x^{p+q} \)[/tex]
### To prove:
[tex]\[ a^{q-r} \times b^{r-p} \times c^{p-q} = 1 \][/tex]
### Step 1: Substitute the expressions for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
#### Substitution for [tex]\( a^{q-r} \)[/tex]:
[tex]\[ a^{q-r} = (x^{q+r} \cdot y^p)^{q-r} \][/tex]
[tex]\[ a^{q-r} = x^{(q+r)(q-r)} \cdot y^{p(q-r)} \][/tex]
[tex]\[ a^{q-r} = x^{q^2 - r^2} \cdot y^{pq - pr} \][/tex]
#### Substitution for [tex]\( b^{r-p} \)[/tex]:
[tex]\[ b^{r-p} = (x^{r+p} \cdot y^q)^{r-p} \][/tex]
[tex]\[ b^{r-p} = x^{(r+p)(r-p)} \cdot y^{q(r-p)} \][/tex]
[tex]\[ b^{r-p} = x^{r^2 - p^2} \cdot y^{qr - qp} \][/tex]
#### Substitution for [tex]\( c^{p-q} \)[/tex]:
[tex]\[ c^{p-q} = (x^{p+q})^{p-q} \][/tex]
[tex]\[ c^{p-q} = x^{(p+q)(p-q)} \][/tex]
[tex]\[ c^{p-q} = x^{p^2 - q^2} \][/tex]
### Step 2: Multiply the expressions [tex]\( a^{q-r} \)[/tex], [tex]\( b^{r-p} \)[/tex], and [tex]\( c^{p-q} \)[/tex]:
[tex]\[ a^{q-r} \times b^{r-p} \times c^{p-q} = (x^{q^2 - r^2} \cdot y^{pq - pr}) \times (x^{r^2 - p^2} \cdot y^{qr - qp}) \times x^{p^2 - q^2} \][/tex]
### Step 3: Combine the exponents for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] separately:
#### For [tex]\( x \)[/tex]:
[tex]\[ x^{(q^2 - r^2) + (r^2 - p^2) + (p^2 - q^2)} \][/tex]
#### For [tex]\( y \)[/tex]:
[tex]\[ y^{(pq - pr) + (qr - qp)} \][/tex]
### Step 4: Simplify the exponents:
#### Simplify the exponent for [tex]\( x \)[/tex]:
[tex]\[ (q^2 - r^2) + (r^2 - p^2) + (p^2 - q^2) = q^2 - r^2 + r^2 - p^2 + p^2 - q^2 = 0 \][/tex]
So,
[tex]\[ x^0 = 1 \][/tex]
#### Simplify the exponent for [tex]\( y \)[/tex]:
[tex]\[ (pq - pr) + (qr - qp) = pq - pr + qr - qp = (pq - qp) + (qr - pr) = 0 \][/tex]
So,
[tex]\[ y^0 = 1 \][/tex]
### Step 5: Combining the simplified expressions:
[tex]\[ a^{q-r} \times b^{r-p} \times c^{p-q} = x^0 \cdot y^0 = 1 \cdot 1 = 1 \][/tex]
Thus, we have successfully proven:
[tex]\[ a^{q-r} \times b^{r-p} \times c^{p-q} = 1 \][/tex]
