Suppose a⋄ba \diamond ba⋄b means ab+1ab + 1ab+1 and a∙ba \bullet ba∙b means a+ba + ba+b. Evaluate4⋄((6∙8)∙(3⋄5)).4 \diamond ((6 \bullet 8) \bullet (3 \diamond 5)).4⋄((6∙8)∙(3⋄5)). Solution 4⋄((6∙8)∙(3⋄5))=4⋄((6+8)∙(3⋅5+1))=4⋄(14∙16)=4⋄(14+16)=4⋄30=4⋅30+1=121.\begin{aligned}4 \diamond ((6 \bullet 8) \bullet (3 \diamond 5)) &= 4 \diamond ((6 + 8) \bullet (3 \cdot 5 + 1))\\&= 4 \diamond (14 \bullet 16)\\&= 4 \diamond (14 + 16)\\&= 4 \diamond 30\\&= 4 \cdot 30 + 1\\&= 121.\end{aligned}4⋄((6∙8)∙(3⋄5))=4⋄((6+8)∙(3⋅5+1))=4⋄(14∙16)=4⋄(14+16)=4⋄30=4⋅30+1=121.