Question

Solve it.....
Mathematics question.​

Answer 1

SOLUTION

GIVEN

$\sf{a = x \: {y}^{p - 1} ,b = x \: {y}^{q - 1} , c = x \: {y}^{r - 1} }$

TO PROVE

$\sf{ {a}^{q - r} \: {b}^{r - p} \: {c}^{p - q} = 1 }$

PROOF

LHS

$\sf{ = {a}^{q - r} \times {b}^{r - p} \times {c}^{p - q} }$

$\sf{ = {(x \: {y}^{p - 1} )}^{q - r} \times {(x \: {y}^{q- 1} )}^{r - p} \times {(x \: {y}^{r- 1} )}^{p - q} }$

$\sf{ = {x}^{q - r + r - p + p - q} \times {y}^{(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)} }$

$\sf{ = {x}^{0} \times {y}^{(pq - pr - q + r) + (qr - pq - r + p) + ( pr -q r- p + q)} }$

$\sf{ = {x}^{0} \times {y}^{0} }$

$= 1\times 1$

$= 1$

= RHS

Hence proved

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Answer 2

The proof is given below:

Given:

$a=xy^{p-1}$

$b=xy^{q-1}$

$c=xy^{r-1}$

To show that:

$a^{q-r}b^{r-p}c^{p-q}=1$

Solution:

$a=xy^{p-1}$

$\implies a^{q-r}=x^{q-r}y^{(p-1)(q-r)}$

$b=xy^{q-1}$

$\implies b^{r-p}=x^{r-p}y^{(q-1)(r-p)}$

$c=xy^{r-1}$

$\implies c^{p-q}=x^{p-q}y^{(r-1)(p-q)}$

Now,

$a^{q-r}b^{r-p}c^{p-q}$

$=x^{q-r}y^{(p-1)(q-r)}\times x^{r-p}y^{(q-1)(r-p)}\times x^{p-q}y^{(r-1)(p-q)}$

$=x^{q-r+r-p+p-q}y^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}$

$=x^0y^{pq-pr-q+r+qr-pq-r+p+pr-qr-p+q}$

$=x^0y^0$

$=1$            ($\because a^0=1$)

Hope this answer is helpful.

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