Question

Factorise it

$64 {}^{3} + 125y {}^{3} - 8z {}^{3} + 120xyz$
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Answer 1

Answer :-

4x+5y+2z).(16x²+25y+4z²-20xy-10yz-8xz )

Solution :-

We have to factories → 64x³ + 125y³ - 2z³ + 120xyz .

→ 64 = 4³ , 125 = 5³ , 8 = 2³

Now replacing these cubes in the question given .

→ 4x³ + 5y³ + 8z³ - 120xyz

We know an important identity ie

$\boxed {{a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca}$

→ Now using this identity .

→ a = 4x , b = 5y , c = 2z

→ 3 a.b.c = 3 × 40 = 120 .

→ ( 4x + 5y + 2z ). ( 16x² + 25y² + 4z² - 20xy - 10yz - 8xz )

So the required answer is ( 4x + 5y + 2z ) ( 16x² + 25y² + 4z² - 20xy - 10yz - 8xz )

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Some more identities :-

→ ( a+b+c )² = a²+b²+c³+2(ab+bc+ca)

→ (a³+b³) = (a+b) × ( a²-ab+b²)

→ (a+b)³ = a³+ b³+3ab(a+b)