Question

if x+y+z=9 find (3-x)^3+(3-y)^3+(3-z)^3-3(3-x)(3-y)(3-z)

Answer 1

Since a3+b3+c3–3abc=(a+b+c)(a2+b2+c2–ab−bc−ac)a3+b3+c3–3abc=(a+b+c)(a2+b2+c2–ab−bc−ac)

In the expression to be evaluated, a=2-x,b=2-y,c=2-z and a+b+c=6-(x+y+z)

Given x+y+z=6, a+b+c=0,

So, a3+b3+c3–3abc=(0)(a2+b2+c2–ab−bc−ac)=0a3+b3+c3–3abc=(0)(a2+b2+c2–ab−bc−ac)=0

If it is known that the answer is constant, you can also rely on checking with 3–4 test cases satisfying given equation such as(2,2,2),(1,2,3),(0,2,4),(0,0,6) etc.

For example:

Let-

2−x=a,2−y=b2−x=a,2−y=band2−z=c2−z=c

Now,

a+b+ca+b+c

=(2−x)+(2−y)+(2−z)=(2−x)+(2−y)+(2−z)

=6−(x+y+z)=6−6=0=6−(x+y+z)=6−6=0

The given expression is-

a3+b3+c3−3abca3+b3+c3−3abc

=(a+b+c)(a2+b2+c2−ab−bc−ac)=(a+b+c)(a2+b2+c2−ab−bc−ac)

=0

For example :

Remember the identity,

a^3+b^3+c^3–3abc=0 if a+b+c=0

Here,a=2-x, b=2-y, c=2-z

So a+b+c= 2-x+2-y+2-z =6-(x+y+z )=6–6=0

(given x+y+z=6)

So the given expression is also equal to zero.

For example :

Since a3+b3+c3–3abc=(a+b+c)(a2+b2+c2–ab−bc−ac)a3+b3+c3–3abc=(a+b+c)(a2+b2+c2–ab−bc−ac)

In the expression to be evaluated, a=2-x,b=2-y,c=2-z and a+b+c=6-(x+y+z)

Given x+y+z=6, a+b+c=0,

So, a3+b3+c3–3abc=(0)(a2+b2+c2–ab−bc−ac)=0a3+b3+c3–3abc=(0)(a2+b2+c2–ab−bc−ac)=0

If it is known that the answer is constant, you can also rely on checking with 3–4 test cases satisfying given equation such as(2,2,2),(1,2,3),(0,2,4),(0,0,6) etc.

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Rhea Saikia, studied Mathematics at University of Delhi (2017)

Answered Jun 22, 2017

A2A

Let-

2−x=a,2−y=b2−x=a,2−y=band2−z=c2−z=c

Now,

a+b+ca+b+c

=(2−x)+(2−y)+(2−z)=(2−x)+(2−y)+(2−z)

=6−(x+y+z)=6−6=0=6−(x+y+z)=6−6=0

The given expression is-

a3+b3+c3−3abca3+b3+c3−3abc

=(a+b+c)(a2+b2+c2−ab−bc−ac)=(a+b+c)(a2+b2+c2−ab−bc−ac)

=0=0

359 Views · View Upvoters · Answer requested by Joytika Kaur

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Nisha Krishnakumar

Answered Jun 22, 2017

Remember the identity,

a^3+b^3+c^3–3abc=0 if a+b+c=0

Here,a=2-x, b=2-y, c=2-z

So a+b+c= 2-x+2-y+2-z =6-(x+y+z )=6–6=0

(given x+y+z=6)

So the given expression is also equal to zero.

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Meet Mehta, former None

Answered Jun 22, 2017

Answer :0

(a^3 +b^3 +c^3 - 3 abc)

=(a+b+c){(a-b)^2 + (b-c) ^2 +(c-a) ^2}

So,

Here (a+b+c)

=(2-x+2-y+2-z)

=6-(a+b+c)

=6–6

=0

Answer 2

(3-x)^3+(3-y)^3+(3-z)^3-3(3-x)(3-y)(3-z) = 0