Question

write in expanded form:i. (2x + 3y)³
ii. (3x-2y)³
iii. (7a+5b)³​

Answer 1

The answer of :

I) 8x³+27y³+18xy(2x+3y)

II) 27x³–8y³–18xy(3x–2y)

III) 343a³+125b³3+126ay(7a+6y)

Step-by-step explanation:

I) (2x+3y)³

→ (2x)³+(3y)³+3•(2x)²•3y + 3•2x•(3y)²

→ 8x³+27y³+3•2x•3y(2x+3y)

→ 8x³+27y³+18xy(2x+3y)

II) (3x-2y)³

→ (3x)³–(2y)³–3•(3x)²•2y + 3•3x•(2y)²

→ 27x³–8y³–3•3x•2y(3x–2y)

→ 27x³–8y³–18xy(3x–2y)

III) (7a+5b)³

→ (7a)³+(5b)³+3•(7a)²•6y+3•7a•(6y)²

→ 343a³+125b³+3•7a•6y(7a+6y)

→ 343a³+125b³3+126ay(7a+6y)

IDENTITIES USED :

(I) and (III) is (a+b)³

(a+b)³ = a³+b³+3ab(a+b)

(II) is (a–b)³

(a–b)³ = a³–b³–3ab(a–b)