Let's simplify step-by-step.2r^2+3s^3−r^2+4t^2−r^2=2^r2+3s^3+−r^2+4t^2+−r^2
Combine Like Terms:=2r^2+3s^3+−r^2+4t^2+−r^2=(3s^3)+(2r^2+−r^2+−r^2)+(4t^2)=3s^3+4t^2
Answer:= 3s^3 + 4t^2
2r2+3s3-r2+4t2-r2
Final result : 3s^3 + 4t^2
(1): "r^2" was replaced by "r^2". 4 more similar replacement(s).
Step by step solution : Step 1 :Equation at the end of step 1 : ((((2•(r^2))+(3•(s^3)))-(r^2))+22t^2)-r^2
Step 2 :Equation at the end of step 2 : ((((2•(r^2))+3s^3)-r^2)+22t^2)-r^2
Step 3 :Equation at the end of step 3 : (((2r^2 + 3s^3) - r^2) + 22t^2) - r^2
Step 4 :Trying to factor as a Sum of Cubes : 4.1 Factoring: 3s^3 + 4t^2
Theory : A sum of two perfect cubes, a^3 + b^3 can be factored into :
(a + b) • (a^2-ab+b^2)
Proof : (a + b) • (a^2-ab+b^2) =
a3-a^2b+ab^2+ba^2-b^2a+b3 =
a^3+(a^2b-ba^2)+(ab^2-b^2a)+b^3=
a^3+0+0+b^3=
a^3+b^3
Check : 3 is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes
Final result : 3s^3 + 4t^2
