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Albert1
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$a+b+c+3=2(\sqrt a +\sqrt {b+1}+\sqrt {c-1})$$find:a^3+b^3+c^3=?$
Albert said:$a+b+c+3=2(\sqrt a +\sqrt {b+1}+\sqrt {c-1})$$find:a^3+b^3+c^3=?$
kaliprasad said:Let a = $x^2$, b = $y^2-1$, c = $z^2+ 1$
We get $ x^2 + y^2 + z^2 + 3 = 2x + 2y + 2z$
Or$ (x^2 – 2x + 1) + (y^2 – 2y + 1) + (z^2-2z +1)=0$
Or $(x-1)^2 + (y-1)^2 + (z-1)^2 = 0$
x = y = z = 1 or a = 1, b = 0, c = 2 => $a^3 + b^3 + c^3$ = 9
The purpose of finding a³ + b³ + c³ is to simplify and represent a polynomial expression in a more concise form. This can be useful in solving equations and identifying patterns in data.
The steps involved in finding a³ + b³ + c³ may vary depending on the context, but generally it involves identifying the variables (a, b, and c), cubing each of them, and then combining them using addition.
Yes, the distributive property can be used to expand a³ + b³ + c³. For example, (a + b + c)(a² - ab + b²) is equal to a³ + b³ + c³.
Finding a³ + b³ + c³ can be applied in various fields such as physics, engineering, and economics. For instance, in physics, it can be used to simplify equations in mechanics and thermodynamics. In economics, it can be used to model demand and supply equations.
Yes, there is a general formula for finding a³ + b³ + c³ which is (a + b + c)(a² - ab + b² - ac - bc + c²). This formula can also be expanded to include more variables, such as (a + b + c + d)(a² - ab + b² - ac - bc + c² - ad - bd - cd + d²).