Question

x^a . x^b . x^c = 1 then prove that-,

x^a^2/bc + x^b^2/ac + x^c^2/ab = x^3

Answer 1

Answer:

$x^a\times x^b \times x^c = 1\\\\\implies x^{a+b+c}=x^0\\\\\implies a+b+c=0$

$x^{\dfrac{a^2}{bc}}\times x^{\dfrac{b^2}{ac}}\times x^{\dfrac{c^2}{ab}}\\\\\implies x^{\dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}}\\\\\implies x^{\dfrac{a^3+b^3+c^3}{abc}}\\\\\implies x^{\dfrac{3abc}{abc}}\\\\\implies x^3$

Step-by-step explanation:

Use the basic fundamentals of indices to start with .

Use $\bf{a^b\times a^c=a^{b+c}}$

Then we use the expansion a³ + b³ + c³ = 3 abc when a + b + c = 0 .

We also know that :

anything raised to the power zero is 1 .

Note that 'anything' should not be a zero .

Also when $x^n=x^m$ then m=n .

Answer 2

ANSWER:-------------

Answer:

xa×xb×xc=1

⟹xa+b+c=x0

⟹a+b+c=0

x^b \times x^c = 1

x^{a+b+c}=x^0

a+b+c=0

{xa+b+c=x0⟹a+b+c=0

xa2bc×xb2

ac×xc2ab

⟹xa2bc+b2ac+c2ab⟹

xa3+b3+c3abc

⟹x3abcabc⟹x3\}

x^{{a^2}{bc}}

x^{{b^2}{ac}}

x^{{c^2}{ab}

x^{{a^2}

{bc}+{b^2}{ac}+\

c^2}{ab}

x^{\{a^3+b^3+c^3}{abc}}

x^{\dfrac{3abc}{abc}}

x^3\end

xbca2×xacb

2×xabc2⟹

xbca2+acb

2+abc2⟹

xabca3+b3+c3⟹

xabc3abc⟹x3

hope it helps:--

T!—!ANKS!!!