3√3a3-b3-5√5c3-3√15abc
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Hi ,
*****************************************
We know the algebraic identity :
x³ + y³ + z³ - 3xyz
= (x+y+z)(x²+y²+z²-xy-yz-zx )
*******************************************
Here ,
x = √3a
y = -b ,
z = -√5c;
Therefore,
3√3a³-b³-5√5c³-3√15abc
= (√3a )³ + ( - b )³ + ( -√5c )-3×√3a×(-b)×(-√5c)
=(√3a-b-√5c)[(√3a)²+(-b)²+(-√5c)²-√3a(-b)-(-b)(-√5c)-(-√5c)(√3a )]
= (√3a-b-√5c)[3a²+b²+5c²+√3ab-√5bc+√15ac]
I hope this helps you.
: )
*****************************************
We know the algebraic identity :
x³ + y³ + z³ - 3xyz
= (x+y+z)(x²+y²+z²-xy-yz-zx )
*******************************************
Here ,
x = √3a
y = -b ,
z = -√5c;
Therefore,
3√3a³-b³-5√5c³-3√15abc
= (√3a )³ + ( - b )³ + ( -√5c )-3×√3a×(-b)×(-√5c)
=(√3a-b-√5c)[(√3a)²+(-b)²+(-√5c)²-√3a(-b)-(-b)(-√5c)-(-√5c)(√3a )]
= (√3a-b-√5c)[3a²+b²+5c²+√3ab-√5bc+√15ac]
I hope this helps you.
: )
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