Question

SOLVE THE FOLLOWING​

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

• Show that :  $\sf{\left(x^{3n+1}\cdot \:x^{3n-1}\right)^2=x^{12n}}$

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♣ ꜰᴏʀᴍᴜʟᴀᴇ ᴜꜱᴇᴅ :

• $\sf{a^m\cdot a^n=a^{m+n}}$

• $\sf{(a^m)^n=a^{mn}}$

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♣ ᴀɴꜱᴡᴇʀ :

$\sf{\left(x^{3n+1}\cdot \:x^{3n-1}\right)^2=x^{12n}}$

Apply Exponential Rule :  $\sf{a^m\cdot a^n=a^{m+n}}$

$\sf{\left(x^{3n+1}\cdot \:x^{3n-1}\right)^2=x^{12n}}$

$\implies\sf{\left(x^{(3n+1)+(3n-1)}\right)^2=x^{12n}}$

$\implies\sf{\left(x^{3n+1+3n-1}\right)^2=x^{12n}}$

Cancelling +1 and -1 :

$\implies\sf{\left(x^{3n+3n}\right)^2=x^{12n}}$

$\implies\sf{\left(x^{6n}\right)^2=x^{12n}}$

Apply Exponential Rule : $\sf{(a^m)^n=a^{mn}}$

$\implies\sf{\left(x\right)^{6n\cdot 2}=x^{12n}}$

$\implies\sf{\left(x\right)^{12n}=x^{12n}}$

$\large\boxed{\implies\sf{x^{12n}=x^{12n}}}$

We showed that $\sf{\left(x^{3n+1}\cdot \:x^{3n-1}\right)^2=x^{12n}}$

Hence Proved !!!