Question

maths ch1 pg 1.36 q28 SM
plz explain in detail step by step ​

Answer 1

$\mathsf{Given :\;\bigg[\dfrac{x^{ab}}{x^{a^2 + b^2}}\bigg]^{a + b} \times \bigg[\dfrac{x^{bc}}{x^{b^2 + c^2}}\bigg]^{b + c} \times \bigg[\dfrac{x^{ca}}{x^{c^2 + a^2}}\bigg]^{c + a}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\dfrac{a^m}{a^n} = a^{m - n}}}}$

$\mathsf{\implies \big[x^{ab - (a^2 + b^2)}\big]^{a + b} \times \big[x^{bc - (b^2 + c^2)}\big]^{b + c} \times \big[x^{ca - (c^2 + a^2)}\big]^{c + a}}$

$\mathsf{\implies \big[x^{-(a^2 + b^2 - ab)}\big]^{a + b} \times \big[x^{-(b^2 + c^2 - bc)}\big]^{b + c} \times \big[x^{-(c^2 + a^2 - ca)}\big]^{c + a}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\big[(a)^m\big]^n = (a)^{mn}}}}$

$\mathsf{\implies \big[x\big]^{-(a^2 + b^2 - ab)(a + b)} \times \big[x\big]^{-(b^2 + c^2 - bc)(b + c)} \times \big[x\big]^{-(c^2 + a^2 - ca)(c + a)}}$

★  We know that : p³ + q³ = (p + q)(p² - pq + q²)

$\mathsf{\implies \big[x\big]^{-(a^3 + b^3)} \times \big[x\big]^{-(b^3 + c^3)} \times \big[x\big]^{-(c^3 + a^3)}}$

$\mathsf{\implies \big[x\big]^{-a^3 - b^3} \times \big[x\big]^{-b^3 - c^3} \times \big[x\big]^{-c^3 - a^3}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{(a)^m \times(a)^n = (a)^{m + n}}}}$

$\mathsf{\implies \big[x\big]^{-a^3 - b^3 - b^3 - c^3 - c^3 - a^3}}$

$\mathsf{\implies \big[x\big]^{-2a^3 - 2b^3 - 2c^3}}$

$\mathsf{\implies \big[x\big]^{-2(a^3 + b^3 + c^3)}}$

Answer 2

[Note : refer the attachment for the solution]

$\boxed{\textbf{\large{step by step explanation}}}$

Step 1 :

First, we have to use identity

( (a^m)/(a^n ))^l

= (a^( m x l )/(a^(n x l) )

From this multiply the terms of power out side the bracket i.e power of the bracket to the numerator power and denominator power

step 2 :

Now, simply apply multiplication of the terms of power

Step 3 :

from the identity property

[( a^m x a^n) = ( a^ (m + n))]

Adding all the terms of powers

step 4 :

Now, As we can see both like terms of numerator powers and denominator powers cancelled out due to the , identity

( a^m / a^n ) = ( a^( m - n)), As if we apply this identity then like terms of power of x are simply cancelled out because of opposite sign.

Step 5 :

Now remaining term is

[( 1 )/ ( (x)^(a^3+b^3+b^3+c^3+c^3+a^3 )]

Then, take the denominator term as at the side of numerator, the sign of power changes due to the identity [ ( 1 / a^(m)) ] = a ^(- m)

Step 6 :

Now do an addition for the like terms of the power x

,then it will form

[ (x) ^(-( 2a + 2b + 2c) )]

Step 7 :

we can place 2 at outside the bracket as a common, negative sign is remains there, so the finally reduced term is -

[ ( x) ^(-2 ( a^3 + b^3 + c^3 ))]

option c is our answer

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for such a type of problems ,we need to use certain identities given below.

◾( m / n) ^2 =( m )^2 / (n) ^2

◾ √( m / n) = √m / √n

◾( a ^(m )^(n)) = ( a^ ( m ✖ n))

◾( a ^( - m)) = 1 /( a ^(m) )

◾( a ^m ✖ a^n ) = ( a ^( m + n))

◾( 1 ) ^(z) = (1)

◾(a^(m)) /(a^(n)) = a^(m - n)

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