If x is a positive real number and exponents are rational numbers, then simplify the following :
(x(a+b))2(x(b+c))2(x(c+a))2(xaxbxc)4\dfrac{(x^{(a + b)})^2(x^{(b + c)})^2(x^{(c + a)})^2}{(x^ax^bx^c)^4}(xaxbxc)4(x(a+b))2(x(b+c))2(x(c+a))2
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Given,
⇒(x(a+b))2(x(b+c))2(x(c+a))2(xaxbxc)4=x2a+2b.x2b+2c.x2c+2ax(a+b+c)4=x2a+2b+2b+2c+2c+2ax4a+4b+4c=x4a+4b+4cx4a+4b+4c=1.\Rightarrow \dfrac{(x^{(a + b)})^2(x^{(b + c)})^2(x^{(c + a)})^2}{(x^ax^bx^c)^4} = \dfrac{x^{2a + 2b}.x^{2b +2c}.x^{2c + 2a}}{x^{(a + b + c)4}} \\[1em] = \dfrac{x^{2a + 2b + 2b + 2c + 2c + 2a}}{x^{4a + 4b + 4c}} \\[1em] = \dfrac{x^{4a + 4b + 4c}}{x^{4a + 4b + 4c}} \\[1em] = 1.⇒(xaxbxc)4(x(a+b))2(x(b+c))2(x(c+a))2=x(a+b+c)4x2a+2b.x2b+2c.x2c+2a=x4a+4b+4cx2a+2b+2b+2c+2c+2a=x4a+4b+4cx4a+4b+4c=1.
Hence, (x(a+b))2(x(b+c))2(x(c+a))2(xaxbxc)4\dfrac{(x^{(a + b)})^2(x^{(b + c)})^2(x^{(c + a)})^2}{(x^ax^bx^c)^4}(xaxbxc)4(x(a+b))2(x(b+c))2(x(c+a))2 = 1.
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If x = am + n, y = an + l and z = al + m, prove that xmynzl = xnylzm.
Show that (p+1q)m×(p−1q)n(q+1p)m×(q−1p)n=(pq)m+n\dfrac{\Big(p + \dfrac{1}{q}\Big)^m \times \Big(p - \dfrac{1}{q}\Big)^n}{\Big(q + \dfrac{1}{p}\Big)^m \times \Big(q - \dfrac{1}{p}\Big)^n} = \Big(\dfrac{p}{q}\Big)^{m + n}(q+p1)m×(q−p1)n(p+q1)m×(p−q1)n=(qp)m+n
(xa2xb2)1a+b(xb2xc2)1b+c(xc2xa2)1c+a\Big(\dfrac{x^{a^2}}{x^{b^2}}\Big)^{\dfrac{1}{a + b}}\Big(\dfrac{x^{b^2}}{x^{c^2}}\Big)^{\dfrac{1}{b + c}}\Big(\dfrac{x^{c^2}}{x^{a^2}}\Big)^{\dfrac{1}{c + a}}(xb2xa2)a+b1(xc2xb2)b+c1(xa2xc2)c+a1
(xbxc)b+c−a(xcxa)c+a−b(xaxb)a+b−c\Big(\dfrac{x^b}{x^c}\Big)^{b + c - a}\Big(\dfrac{x^c}{x^a}\Big)^{c + a - b}\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c}(xcxb)b+c−a(xaxc)c+a−b(xbxa)a+b−c
