(i) 108a2 – 3(b – c)2
(ii) πa5 – π3ab2
Answer :
(i) 108a2 – 3(b – c)2
Take out common in all terms,
3[36a2 – (b – c)2]
3[(6a)2 – (b – c)2]
We know that, a2 – b2 = (a + b) (a – b)
3[(6a + b – c) (6a – b + c)]
(ii) πa5 – π3ab2
πa5 – π3ab2
Take out common in all terms,
πa(a4 – π2b2)
πa((a2)2 – (πb)2)
We know that, a2 – b2 = (a + b) (a – b)
πa(a2 + πb) (a2 – πb)
More Solutions: