Q7 of 19 Page 59

E×pand.

(1) (5 ×-7) (5 ×-9)


(2) (2 ×-3 y)3


(3) (a + 1/2)3


(1) (5×-7)(5×-9)


⇒ 5×(5×-9) – 7(5×-9)


⇒ 25×2 – 45× - 35× + 63


⇒ 25×2 - 80× + 63


(2) (2× – 3y)3


Using identity: (a-b)3 = a3 – b3 – 3a2b – 3ab2 , we get


(2×)3 - (3y)3 – 3(2×)2(3y) + 3(2×)(3y)2


= 8×3 – 27 y3 – 36×2 y + 54×y2


(3) (a + 1/2)3


Using identity : (a+b)3 = a3 + b3 + 3a2b + 3ab2 , we get


(a)3 + 3 + 3(a)2()+ 3(a)()2


= a3 + + a2 + a


More from this chapter

All 19 →
5

Draw a line l. Draw a line parallel to line l at a distance of 3.5 cm.

 6

Fill in the blanks in the following statement.

The number (256)5/7 is .......... th root of .......... th power of .........


 8

Draw an obtuse angled triangle. Draw all of its medians and show their point of concurrence.

 9

Draw ∆ABC such that l(BC) = 5.5 cm, m∠ABC = 90°, l(AB) = 4 cm. Show the orthocentre of the triangle.