If a = x^m + n.y^l; b = x^n + l.y^m and c = x^l + m.y^n, prove that : a^m - n.b^n - l.c^l - m = 1

Substituting values of a, b and c in L.H.S. of equation a m - n .b n - l .c l - m = 1, we get : ⇒ a m - n .b n - l .c l - m = (x m + n .y l ) m - n .(x n + l .y m ) n - l .(x l + m .y n ) l- m = (x (m + n)(m - n) .y l(m - n) ).(x (n + l)(n - l) .y m(n - l) ).(x (l + m)(l - m) .y n(l - m) ) = (x m 2 - n 2 ).(x n 2 - l 2 ).(x l 2 - m 2 ).(y lm - ln ).(y mn - ml ).(y nl - nm ) = x m 2 - n 2 + n 2 - l 2 + l 2 - m 2 .y lm - ln + mn - ml + nl - nm = x 0 .y 0 = 1.1 = 1. Since, L.H.S. = R.H.S. = 1. Hence, proved that a m - n .b n - l .c l- m = 1.

Mathematics

If a = x^{m + n}.y^l; b = x^{n + l}.y^m and c = x^{l + m}.yⁿ,
prove that : a^{m - n}.b^{n - l}.c^{l - m} = 1

Indices

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Answer

Substituting values of a, b and c in L.H.S. of equation a^{m - n}.b^{n - l}.c^{l - m} = 1, we get :

⇒ a^{m - n}.b^{n - l}.c^{l - m} = (x^{m + n}.y^l)^{m - n}.(x^{n + l}.y^m)^{n - l}.(x^{l + m}.yⁿ)^{l- m}

= (x^{(m + n)(m - n)}.y^{l(m - n)}).(x^{(n + l)(n - l)}.y^{m(n - l)}).(x^{(l + m)(l - m)}.y^{n(l - m)})

= (x^{m² - n²}).(x^{n² - l²}).(x^{l² - m²}).(y^{lm - ln}).(y^{mn - ml}).(y^{nl - nm})

= x^{m² - n² + n² - l² + l² - m²}.y^{lm - ln + mn - ml + nl - nm}

= x⁰.y⁰

= 1.1

= 1.

Since, L.H.S. = R.H.S. = 1.

Hence, proved that a^{m - n}.b^{n - l}.c^{l- m} = 1.

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Mathematics

If a = x^{m + n}.y^l; b = x^{n + l}.y^m and c = x^{l + m}.yⁿ,
prove that : a^{m - n}.b^{n - l}.c^{l - m} = 1

Indices

Answer

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$\dfrac{3 \times 27^{n + 1} + 9 \times 3^{3n - 1}}{8 \times 3^{3n} - 5 \times 27^n}$

Show that :
$\Big(\dfrac{a^m}{a^{-n}}\Big)^{m - n} \times \Big(\dfrac{a^n}{a^{-l}}\Big)^{n - l} \times (\dfrac{a^l}{a^{-m}}\Big)^{l - m}$ = 1

Simplify :
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Simplify :
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