Answer :

tejaspadhi12
a³+27b³+8c³-18abc
=a³+(3b)³+(2c)³-3×a×(3b)×(2c)
=x³+y³+z³-3xyz, where a=x, 3b=y, 2c=z
=(x+y+z)(x²+y²+z²-xy-yz-xz)
=(a+3b+2c)(a²+9b²+4c²-3ab-6bc-2ca)
                                                      [Ans]
birendrak1975

a³+27b³+8c³-18abc

(a)³+(3b)³+(2c)³-3xax3bx2c

We know the formula that a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)

where a=a,b=3b,c=3c

(a+3b+2c){(a)²+(3b)²+(2c)²-ax3b-3bx2c-2cxa)

(a+3b+2c)(a²+9b²+4c²-3ab-6bc-2ca)     ..........................(ANS)

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