Factorise the following!
☺ Solve it☺
☺ Needed Explanation ☺
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Answered by
99
Given :
- 24a³ + 81b³
- y³ + 1/8y³
- a³ + 8/a³
- 1 + q³/125
Solution
Apply identity : a³ + b³ = (a + b)(a² - ab + b²)
→ 24a³ + 81b³
Take 3 as a common
→ 3[8a³ + 27b³]
→ 3[(2a)³ + (3b)³]
→ 3[{2a + 3b}{(2a)² - 2a*3b + (3b)²]
→ 3[{2a + 3b}{4a² - 6ab + 9b²}]
- y³ + 1/8y³
Apply identity : a³ + b³ = (a + b)(a² - ab + b²)
→ (y)³ + (1/2y)³
→ (y + 1/2y)[(y)² - y*(1/2y) + (1/2y)²]
→ (y + 1/2)[y² - 1/2 + 1/4y²]
- a³ + 8/a³
Apply identity : a³ + b³ = (a + b)(a² - ab + b²)
→ (a)³ + (2/a)³
→ (a + 2/a)[(a)² - a*2/a + (2/a)²]
→ (a + 2/a)[a² - 2 + 4/a²]
- 1 + q³/125
Apply identity : a³ + b³ = (a + b)(a² - ab + b²)
→ (1)³ + (q/5)³
→ (1 + q/5)[(1)² - 1*q/5 + (q/5)²]
→ (1 + q/5)[(1 - q/5 + q²/25]
Answered by
69
Qᴜᴇsᴛɪᴏɴ ❺ :
➥ 24a³ + 81b³ y² + 1/8³
Sᴏʟᴜᴛɪᴏɴ :
3(8a³ + 27b³)
3[2(a)³ + (3b)³]
3(2a + 3b) {(2a)² + (3b)² - 2a*3b}
Qᴜᴇsᴛɪᴏɴ ❻ :
➥ y³ +
Sᴏʟᴜᴛɪᴏɴ :
Qᴜᴇsᴛɪᴏɴ ❼ :
➥ a³ +
Sᴏʟᴜᴛɪᴏɴ :
a³ +
Qᴜᴇsᴛɪᴏɴ ❽ :
➥ 1 +
Sᴏʟᴜᴛɪᴏɴ :
1 +
Anonymous:
Perfect :P
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