Question

if a=x^m+n.y^l;b=x^n+1.y^m and c=x^l+m.y^n Prove that : a^m-n.b^n-l.c^l-m=1​

Answer 1

GIVEN:

$a= x^{m+n}y^l$

$b= x^{n + l}y^m$

$c= x^{l + m}y^n$

TO PROVE:

$a^{m-n}\times b^{n-l}\times c^{l-m}=1$

EXPLANATION:

$a^{m-n}=\dfrac{a^m}{a^n}$

$a^{m - n}= \dfrac{\left(x^{m+n} \times y^l\right)^m}{\left(x^{m+n} \times y^l\right)^n}$

$a^{m - n}= \dfrac{x^{m(m+n)} \times y^{lm}}{x^{n(m+n)} \times y^{ln}}$

$a^{m - n}= x^{m(m+n) - n(m + n)} \times y^{lm - ln}$

$a^{m - n}= x^{(m+n) (m- n)} \times y^{lm - ln}$

$a^{m - n}= x^{ {m}^{2} - {n}^{2} } \times y^{lm - ln}$

$b^{n - l}=\dfrac{b^n}{b^l}$

$b^{n - l}= \dfrac{\left(x^{n + l} \times y^m\right)^n}{\left(x^{n + l} \times y^m\right)^l}$

$b^{n - l}= \dfrac{x^{n(n + l)} \times y^{mn}}{x^{l(n + l)} \times y^{ml}}$

$b^{n - l}=x^{n(n + l) - l(n + l)} \times y^{mn - ml}$

$b^{n - l}=x^{(n + l) (n- l)} \times y^{mn - ml}$

$b^{n - l}=x^{ {n}^{2} - {l}^{2} } \times y^{mn - ml}$

$c^{l-m} = \dfrac{ {c}^{l} }{ {c}^{m} }$

$c^{l-m} = \dfrac{( x^{l + m} \times y^n)^{l} }{ (x^{l + m} \times y^n) ^{m} }$

$c^{l-m} = \dfrac{ x^{l(l + m)} \times y^{nl }}{ x^{m(l + m)} \times y^{mn} }$

$c^{l-m} = x^{l(l + m) - m(l + m)} \times y^{nl - mn}$

$c^{l-m} = x^{(l + m)(l - m)} \times y^{nl - mn}$

$c^{l-m} = x^{ {l}^{2} - {m}^{2} } \times y^{nl - mn}$

$a^{m-n}\times b^{n-l}\times c^{l-m}=x^{0} \times y^{0} = 1$

Since on multiplying the terms x is common in all the three terms and so we can add the powers.

$x^{{m}^{2} - {n}^{2} + {n}^{2} - {l}^{2} + {l}^{2} - {m}^{2} } = {x}^{0} = 1$

Again on multiplying the terms y is common in all the three terms and so we can add the powers.

$y^{lm - ln + mn - ml +nl - mn } = {y}^{0} = 1$

HENCE PROVED.

Answer 2

$\rm \huge { \underline{ \underline{ \pink{answer}}}}$

$\bf \huge {given}$

$\bf { \huge\star \: a = {x}^{m + n} {y}^{l} }$

$\bf{ \huge \star \: b = {x}^{n + l} {y}^{m} }$

$\bf { \huge \star \: c = {x}^{l + m} {y}^{n} }$

$\bf { \underline{ \underline{ \green{putting \: the \: values \: of \: a \: b \: and \: c \: in \: {a}^{m - n} {b}^{n - l} {c}^{l - m} \: we \: get}}}}$

$\implies\bf \red{ {a}^{m - n} {b}^{n - l} {c}^{l - m} }$

$\implies \bf \red{ {(x}^{m + n} {y}^{l} ) ^{m - n} {(x}^{n + l} {y}^{m} ) ^{n - l} {(x}^{l + m} {y}^{n} ) ^{l - m} }$

$\implies \bf \red{[ {x}^{(m + n)(m - n)} {y}(l(m - n))][ {x}^{(n + l)(n - l) } {y}^{m(n - l)} ][ {x}^{(l + m)(lm)} {y}^{n(l - m)} ]}$

$\implies\bf \red{ {x}^{( {m}^{2} - {n}^{2} ) }{x}^{( {n}^{2} - {l}^{2}) } {x}^{( {l}^{2} - {m}^{2} ) } {y}^{lm - ln} {y}^{mn - ml} {y}^{nl - nm} }$

$\implies \bf \red{ {x}^{0} {y}^{0} }$

$\implies \bf \red{1}$

$\rm \huge { \underline { \underline{ \blue{hence \: proved}}}}$