Question

factorise 125a³+b³+64c³-60abc​

Answer 1

GIVEn

factorise 125a³+b³+64c³-60abc

SOLUTIOn

→ 125a³ + b³ + 64c³ - 60abc

→ (5a)³ + (b)³ + (4c)³ - 3 × 5a × b × 4c

★ Applying identity

a³ + b³ + c³ - 3abc

= (a + b + c)(a² + b² + c² - ab - bc - ca)

→ (5a + b + 4c)[(5a)² + (b)² + (4c)² - 5a*b - b*4c - 5a*4c]

→ (5a + b + 4c)[25a² + b² + 16c² - 5ab - 4cb - 20ac]

SOMe IDENTITIEs

• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab
• a² - b² = (a + b)(a - b)

Answer 2

Fᴀᴄᴛᴏʀɪsᴇ :- ⇩

➠ 125a³ + b³ + 64c³ - 60abc

➠ (5a) ³ + (b) ³ + (4c) ³ - 3 × 5a × b × 4c

Now,

We know that,

➱ a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² -ab - ac - bc)

So,

By applying this identity on the above, we get,

➠ 125a³ + b³ + 64c³ - 60abc = ( 5a + b + 4c) {(5a) ² + (b) ² + (4c) ² -(5a × b) - (5a× 4c) -(b × 4c)

➠ 125a³ + b³ + 64c³ - 60abc = ( 5a + b + 4c) (25a² + b² + 16c² - 5ab - 20ac - 4bc)

➲ ᴹᵒʳᵉ ᴵᵈᵉⁿᵗⁱᵗⁱᵉˢ :-

• (a³ + b³) = (a + b) (a² + b² - ab)

• (a³ - b³) = ( a - b) ( a² + b² + ab)

• If ( a + b + c) = 0 then, a³ + b³ + c³ = 3abc