Answer :

**(b)**

First Part,

Given, (i) x^{4} – 4x^{3} + 6x^{2}

(ii) x^{2}

(i) × (ii)

= (x^{4} – 4x^{3} + 6x^{2}) × x^{2}

= x^{4}x^{2} – 4x^{3}x^{2} + 6x^{2}x^{2}

We know (x^{a}x^{b}= x^{a+b})

= x^{4+2}– 4x^{3+2} + 6x^{2+2}

= x^{6} – 4x^{5} + 6x^{4}

(i) ÷ (ii)

We know

= x^{4-2} – 4x^{3-2} + 6x^{2-2}

= x^{2} – 4x + 6

Second Part

Given, (i) 3m^{2}n^{3} + 40m^{3}n^{4} – 5m^{4}n^{5}

(ii) 10m^{2}n^{2}

(i) × (ii)

= (3m^{2}n^{3} + 40m^{3}n^{4} – 5m^{4}n^{5}) × 10m^{2}n^{2}

We know (x^{a}x^{b}= x^{a+b})

= 30m^{2}m^{2}n^{3}n^{2} + 400m^{3}m^{2}n^{4}n^{2} – 50m^{4}m^{2}n^{5}n^{2}

= 30m^{2+2}n^{3+2} + 400m^{3+2}n^{4+2} – 50m^{4+2}n^{5+2}

= 30m^{4}n^{5} + 400m^{5}n^{6} – 50m^{6}n^{7}

(i) ÷ (ii)

We know

**(c)**

Given,

(i) 49l^{2} – 100m^{2}

(ii) (7l + 10m)

(i) × (ii)

= (49l^{2} – 100m^{2})(7l + 10m)

We know (x^{a}x^{b}= x^{a+b})

= 343l^{2}l+ 490l^{2}m – 700m^{2}l – 1000m^{2}m

= 343l^{2+1} + 490l^{2}m – 700lm^{2} – 1000m^{2+1}

= 343l^{3} + 490l^{2}m – 700lm^{2} – 1000m^{3}

(i) ÷ (ii)

∵ a^{2} – b^{2} = (a – b)(a + b), taking a = 7l, b = 10m we have

Cancelling out the common terms from numerator and denominator, we get

= 7l – 10m

**(d)**

Given,

(i) 625a^{4} – 81b^{4}

(ii) 5a + 3b

(i) × (ii)

= (625a^{4} – 81b^{4})(5a + 3b)

= 625a^{4}a + 1875a^{4}b – 405ab^{4} – 243b^{4}b

We know (x^{a}x^{b}= x^{a+b})

= 625a^{4+1} + 1875a^{4}b – 405ab^{4} – 243b^{4+1}

= 3125a^{5} + 1875a^{4}b – 405ab^{4} – 243b^{5}

(i) ÷ (ii)

∵ a^{2} – b^{2} = (a – b)(a + b), taking a = 25a^{2}, b = 9b^{2}, we have

∵ a^{2} – b^{2} = (a – b)(a + b), taking a = 5a, b = 3b we have

*=*

Cancelling out the common terms from numerator and denominator, we get

= (5a – 3b)(25a^{2} + 9b^{2})

= 125aa^{2} + 45ab^{2} – 75a^{2}b – 27bb^{2}

We know (x^{a}x^{b}= x^{a+b})

= 125a^{2+1} + 45ab^{2} – 75a^{2}b – 27b^{2+1}

= 125a^{3} + 45ab^{2} – 75a^{2}b – 27b^{3}

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