Question

(27)
${27}^3 + {125}^3 = {?}$
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Answer 1

Solution:

27y³ + 125z³

☞ (3y)³ + (5z)³

☞ (3y + 5z)[(3y)² - 3y.5z + (5z)² ]

☞ (3y + 5z) [9y² - 15yz + 25z² ]

Identity Used for this Question:

a³ + b³ = ( a + b ) ( a² - ab + b² )

Important Identities :

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

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

Answer 2

$HELLO GUYS$

QUESTIONS :-

${27}^3 + {125}^3 = {?}$

ANSWER :-

Answer:

27x3−125y3=(3x−5y)(9x2+15xy+25y2)

Explanation:

Notice that both of the terms are perfect cubes:

27x3=(3x)3

125y3=(5y)3

So we can conveniently use the difference of cubes identity:

A3−B3=(A−B)(A2+AB+B2)

with A=3x and B=5y as follows:

27x3−125y3

=(3x)3−(5y)3

=(3x−5y)((3x)2+(3x)(5y)+(5y)2)

=(3x−5y)(9x2+15xy+25y2)

The remaining quadratic factor has no linear factors with Real coefficients. If we allow Complex coefficients then we can factor a little further:

=(3x−5y)(3x−5ωy)(3x−5ω2y)

where ω=−12+√32i is the primitive Complex cube root of 1.